Class 11 - Maths - Binomial Theorem

**Exercise 8.1**

Question 1:

Expand the expression (1– 2x)^{5}

Answer:

By using Binomial Theorem, the expression (1– 2x)^{5} can be expanded as

(1– 2x)^{5} = ^{5}C_{0} (1)^{5} - ^{5}C_{1} (1)^{4} (2x) + ^{5}C_{2} (1)^{3} (2x)^{2} - ^{5}C_{4} (1)^{1} (2x)^{4} + ^{5}C_{5} (2x)^{5}

= 1 – 5(2x) + 10(4x^{2}) – 10(8x^{3}) + 5(16x^{4}) – 32x^{5}

= 1 – 10x + 40x^{2} – 80x^{3} + 80x^{4} – 32x^{5}

Question 2:

Expand the expression (2/x – x/2)^{5}

Answer:

By using Binomial Theorem, the expression (2/x – x/2)^{5} can be expanded as

(2/x– x/2)^{5} = ^{5}C_{0} (2/x)^{5} - ^{5}C_{1} (2/x)^{4} (x/2) + ^{5}C_{2} (2/x)^{3} (x/2)^{2} - ^{5}C_{4} (2/x)^{1} (x/2)^{4} + ^{5}C_{5} (x/2)^{5}

= 32/x^{5} – 5(16/x^{4})(x/2) + 10(8/x^{3})(x^{2}/4) – 10(4/x^{2})(x^{3}/8) + 5(2/x)(x^{4}/16) – x^{5}/32

= 32/x^{5} – 40/x^{3} + 20/x – 5x + 5x^{3}/8 - x^{5}/32

Question 3:

Expand the expression (2x – 3)^{6}

Answer:

By using Binomial Theorem, the expression can be expanded as

(2x – 3)^{6} = ^{6}C_{0} (2x)^{6} - ^{6}C_{1} (2x)^{5} (3) + ^{6}C_{2} (2x)^{4} (3)^{2} - ^{6}C_{3} (2x)^{3} (3)^{2} + ^{6}C_{4} (2x)^{2} (3)^{4} - ^{6}C_{5} (2x)^{1} (3)^{5} +

^{6}C_{6} (3)^{6}

= 64x^{6} – 6(32x^{5})* 3 + 15(16x^{4}) * 9 – 20(8x^{3}) * 27 + 15(4x^{2}) * 81 – 6 * 2x * 243 + 729

= 64x^{6} – 576x^{5} + 2160x^{4} – 4320x^{3} + 4860x^{2} – 2916x + 729

Question 4:

Expand the expression (x/3 + 1/x)^{5}

Answer:

By using Binomial Theorem, the expression (2/x – x/2)^{5} can be expanded as

(x/3 + 1/x)^{5} = ^{5}C_{0} (x/3)^{5} + ^{5}C_{1} (x/3)^{4} (1/x) + ^{5}C_{2} (x/3)^{3} (1/x)^{2} + ^{5}C_{4} (x/3)^{1} (1/x)^{4} + ^{5}C_{5} (1/x)^{5}

= x^{5}/343 + 5(x^{4}/81)(1/x) + 10(x^{3}/27)(1/x^{2}) + 10(x^{2}/9)(1/x^{3}) + 5(x/3)(1/x^{4}) + 1/x^{5}

= x^{5}/343 + 5x^{3}/81 + 10x/27 +10/9x + 5/3x^{3} + 1/x^{5}

Question 5:

Expand the expression (2/x – x/2)^{5}

Answer:

By using Binomial Theorem, the expression (x + 1/x)^{6} can be expanded as

(x + 1/x)^{6} = ^{6}C_{0} (x)^{6} + ^{6}C_{1} (x)^{5} (1/x) + ^{6}C_{2} (x)^{4} (1/x)^{2} + ^{6}C_{3} (x)^{3} (1/x)^{4} + ^{6}C_{4} (x)^{2} (1/x)^{4} + ^{6}C_{5} (x) (1/x)^{5}

+ ^{6}C_{6} (1/x)^{6}

= x^{6} + 6(x)^{5}(1/x) + 14(x)^{4}(1/x^{2}) + 20(x)^{3}(1/x^{3}) + 15(x)^{2}(1/x^{4}) + 6(x)(1/x^{5}) + 1/x^{6}

= x^{6} + 6x^{4} + 15x^{2} + 20 + 15/x^{2} + 6/x^{4} + 1/x^{6}

Question 6:

Using Binomial Theorem, evaluate (96)^{3}

Answer:

96 can be expressed as the sum or difference of two numbers whose powers are easier to

calculate and then, binomial theorem can be applied.

It can be written that, 96 = 100 – 4

Now, (96)^{3} = (100 - 4)^{3}

= ^{3}C_{0} (100)^{3} - ^{3}C_{1} (100)^{2} (4) + ^{3}C_{2} (100)(4)^{2} - ^{3}C_{3} (4)^{3}

= (100)^{3} - 3(100)^{2} (4) + 3(100)(4)^{2} - (4)^{3}

= 1000000 – 120000 + 4800 – 64

= 884736

Question 7:

Using Binomial Theorem, evaluate (102)^{5}

Answer:

102 can be expressed as the sum or difference of two numbers whose powers are easier

to calculate and then, Binomial Theorem can be applied.

It can be written that, 102 = 100 + 2

Now, (102)^{5} = (100 + 2)^{5}

= ^{5}C_{0} (100)^{5} + ^{5}C_{1} (100)^{4} (2) + ^{5}C_{2} (100)^{3}(2)^{2} + ^{5}C_{3} (100)^{2} (2)^{3} + ^{5}C_{4} (100)(2)^{4} + ^{5}C_{5} (2)^{5}

= (100)^{5} + 5(100)^{4} (2) + 10(100)^{3}(2)^{2} + 10(100)^{2}(2)^{3} + 5(100)(2)^{4} + (2)^{5}

= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32

= 11040808032

Question 8:

Using Binomial Theorem, evaluate (101)^{4}

Answer:

101 can be expressed as the sum or difference of two numbers whose powers are easier

to calculate and then, Binomial Theorem can be applied.

It can be written that, 101 = 100 + 1

Now, (101)^{4} = (100 + 1)^{4}

= ^{4}C_{0} (100)^{4} + ^{4}C_{1} (100)^{3} (1) + ^{4}C_{2} (100)^{2}(1)^{2} + ^{4}C_{3} (100) (1)^{3} + ^{4}C_{4} (1)^{4}

= (100)^{4} + 4(100)^{3} + 6(100)^{2} + 4(100)^{ }+ 1

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

Question 9:

Using Binomial Theorem, evaluate (99)^{5}

Answer:

99 can be written as the sum or difference of two numbers whose powers are easier to

calculate and then, Binomial Theorem can be applied.

It can be written that, 99 = 100 – 1

Now, (99)^{5} = (100 - 1)^{5}

= ^{5}C_{0} (100)^{5} - ^{5}C_{1} (100)^{4} (1) + ^{5}C_{2} (100)^{3}(1)^{2} - ^{5}C_{3} (100)^{2} (1)^{3} + ^{5}C_{4} (100)(1)^{4} - ^{5}C_{5} (1)^{5}

= (100)^{5} - 5(100)^{4} + 10(100)^{3} - 10(100)^{2} + 5(100) - 1

= 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1

= 9509900499

Question 10:

Using Binomial Theorem, indicate which number is larger (1.1)^{10000} or 1000.

Answer:

By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)^{10000} can be

obtained as

(1.1)^{10000} = (1 + 0.1)^{10000}

= ^{10000}C_{0} - ^{10000}C_{1} (1.1) + other positive terms

= 1 + 10000 * 1.1 + other positive terms

= 1 + 11000 + other positive terms > 1000

Hence, (1.1)^{10000} > 1000

Question 11:

Find (a + b)^{4} – (a – b)^{4}. Hence, evaluate. (√3 + √2)^{4} – (√3 – √2)^{4}

Answer:

Using Binomial Theorem, the expressions, (a + b)^{4} – (a – b)^{4} can be expanded as

(a + b)^{4} = ^{4}C_{0} a^{4} + ^{4}C_{1} a^{3} b + ^{4}C_{2} a^{2} b^{2} + ^{4}C_{3} a b^{3} + ^{4}C_{4} b^{4}

(a - b)^{4} = ^{4}C_{0} a^{4} - ^{4}C_{1} a^{3} b + ^{4}C_{2} a^{2} b^{2} - ^{4}C_{3} a b^{3} + ^{4}C_{4} b^{4}

Now, (a + b)^{4} – (a – b)^{4} = [^{4}C_{0} a^{4} + ^{4}C_{1} a^{3} b + ^{4}C_{2} a^{2} b^{2} + ^{4}C_{3} a b^{3} + ^{4}C_{4} b^{4}] – [^{4}C_{0} a^{4} - ^{4}C_{1} a^{3} b + ^{4}C_{2} a^{2}

b^{2} - ^{4}C_{3} a b^{3} + ^{4}C_{4} b^{4}]

= 2[^{4}C_{1} a^{3} b + ^{4}C_{3} a b^{3}]

= 8ab(a^{2} + b^{2})

Now, put a = √3 and b = √2, we get

(√3 + √2)^{4} – (√3 – √2)^{4} = 8(√3)( √2){( √3)^{2} + (√2)^{2}}

= 8(√6)(3 + 2)

= 40√6

Question 12:

Find (x + 1)^{6} + (x – 1)^{6}. Hence or otherwise evaluate (√2 + 1)^{6} + (√2 – 1)^{6}

Answer:

Using Binomial Theorem, the expressions, (x + 1)^{6} + (x – 1)^{6} can be expanded as

(x + 1)^{6} = ^{6}C_{0} x^{6} + ^{6}C_{1} x^{5} + ^{6}C_{2} x^{4} + ^{6}C_{3} x^{3} + ^{6}C_{4} x^{2} + ^{6}C_{5} x + ^{6}C_{6}

(x - 1)^{6} = ^{6}C_{0} x^{6} - ^{6}C_{1} x^{5} + ^{6}C_{2} x^{4} - ^{6}C_{3} x^{3} + ^{6}C_{4} x^{2} - ^{6}C_{5} x + ^{6}C_{6}

Now, (x + 1)^{6} + (x – 1)^{6} = 2[^{6}C_{0} x^{6} + ^{6}C_{2} x^{4} + ^{6}C_{4} x^{2} + ^{6}C_{6}]

= 2[x^{6} + 15x^{4} + 15x^{2} + 1]

By putting x = √2, we obtain

(√2 + 1)^{6} + (√2 – 1)^{6} = 2[(√2)^{6} + 15(√2)^{4} + 15(√2)^{2} + 1]

= 2(8 + 15 * 4 + 15 * 2 + 1)

= 2(8 + 60 + 30 + 1)

= 2 * 99

= 198

Question 13:

Show that 9^{n+1} - 8n - 9 is divisible by 64, whenever n is a positive integer.

Answer:

In order to show that 9^{n+1} - 8n - 9 is divisible by 64, it has to be proved that,

9^{n+1} - 8n – 9 = 64k, where k is some natural number

By Binomial Theorem,

(1 + a)^{m} = ^{m}C_{0} + ^{m}C_{1} a + ^{m}C_{2} a^{2} +………..+ ^{m}C_{m} a^{m}

For a = 8 and m = n + 1, we obtain

(1 + 8)^{n+1} = ^{n+1}C_{0} + ^{n+1}C_{1} 8 + ^{n+1}C_{2} 8^{2} +………..+ ^{n+1}C_{n+1} 8^{n+1}

=> 9^{n+1} = 1 + 8(n + 1) + 8^{2} [^{n+1}C_{2} + ^{n+1}C_{3} 8 +………..+ ^{n+1}C_{n+1} 8^{n-1}]

=> 9^{n+1} = 1 + 8n + 8 + 64[^{n+1}C_{2} + ^{n+1}C_{3} 8 +………..+ ^{n+1}C_{n+1} 8^{n-1}]

=> 9^{n+1} = 9 + 8n + 64[^{n+1}C_{2} + ^{n+1}C_{3} 8 +………..+ ^{n+1}C_{n+1} 8^{n-1}]

=> 9^{n+1} - 8n – 9 = 64[^{n+1}C_{2} + ^{n+1}C_{3} 8 +………..+ ^{n+1}C_{n+1} 8^{n-1}]

=> 9^{n+1} - 8n – 9 = 64k, where k = [^{n+1}C_{2} + ^{n+1}C_{3} 8 +………..+ ^{n+1}C_{n+1} 8^{n-1}], is a natural number.

Hence, 9^{n+1} - 8n - 9 is divisible by 64, whenever n is a positive integer.

Question 14:

Prove that: ^{r n}C_{r} = 4^{n}

Answer:

By Binomial Theorem,

^{ n}C_{r} a^{n-r} b^{r} = (a + b)^{n}

By putting b = 3 and a = 1 in the above equation, we obtain

^{ n}C_{r} 1^{n-r} 3^{r} = (1 + 3)^{n}

=> ^{r n}C_{r} = 4^{n}

Hence, proved.

**Exercise 8.2**

Question 1:

Find the coefficient of x^{5} in (x + 3)^{8}

Answer:

It is known that (r + 1)^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Assuming that x^{5} occurs in the (r + 1)^{th} term of the expansion (x + 3)^{8}, we obtain

T_{8+1} = ^{8}C_{r} a^{8 - r} 3^{r}

Comparing the indices of x in x^{5} and in T_{r +1},we obtain r = 3

Thus, the coefficient of x^{5} is

^{8}C_{3} 3^{3} = 8!/(3! * 5!) * 3^{3} = (8 * 7 * 6 * 5!)/(3 * 2 * 1 * 5!) * 3^{3}

= 8 * 7 * 27

= 1512

Question 2:

Find the coefficient of a^{5} b^{7} in (a – 2b)^{12}

Answer:

It is known that (r + 1)^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Assuming that a^{5} b^{7} occurs in the (r + 1)^{th} term of the expansion (a – 2b)^{12}, we obtain

T_{8+1} = ^{12}C_{r} a^{12 - r} (-2b)^{r }= ^{12}C_{r} (-2)^{r} a^{12 - r} (b)^{r }

Comparing the indices of x in a^{5} b^{7} and in T_{r +1}, we obtain r = 7

Thus, the coefficient of a^{5} b^{7} is

^{12}C_{7} (-2)^{7} = -12!/(7! * 5!) * 2^{7} = -(12 * 11 * 10 * 9 * 8 * 7!)/(5 * 4 * 3 * 2 * 1 * 5!) * 2^{7}

= -792 * 128

= -101376

Question 3:

Write the general term in the expansion of (x^{2} – y)^{6 }

Answer:

It is known that the general term T_{r+1} {which is the (r + 1)^{th} term} in the binomial expansion of

(a + b)^{n} is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}.

Thus, the general term in the expansion of (x^{2} – y)^{6 }is

T_{r +1} = ^{6}C_{r} (x^{2})^{6 - r} (-y)^{r }= (-1)^{r} ^{6}C_{r} c^{12 -2 r} y^{r }

Question 4:

Write the general term in the expansion of (x^{2} – yx)^{12}, x ≠ 0

Answer:

It is known that the general term Tr+1 {which is the (r + 1)^{th} term} in the binomial expansion

of (a + b)^{n} is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}.

Thus, the general term in the expansion of (x^{2} – yx)^{12} is

T_{r +1} = ^{12}C_{r} (x^{2})^{12 - r} (-yx)^{r }= (-1)^{r} ^{12}C_{r} x^{24 -2r} y^{r }x^{r} = (-1)^{r} ^{12}C_{r} x^{24 - r} y^{r}

Question 5:

Find the 4^{th} term in the expansion of (x – 2y)^{12}.

Answer:

It is known that (r + 1)^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Thus, the 4^{th} term in the expansion of (x – 2y)^{12} is

T_{4} = T_{3 +1} = ^{12}C_{3} (x^{2})^{12 - 3} (-2y)^{3 }= (-1)^{3} * 12!/(9! * 3!) x^{9} 2^{3 }y^{3}

= -(12 * 11 * 10)/(3 * 2) * 2^{3} * x^{9} y^{3}

= -1760 x^{9} y^{3}

Question 6:

Find the 13^{th} term in the expansion of (9x – 1/3√x)^{18}, x ≠ 0

Answer:

^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Thus, the 13^{th} term in the expansion of (9x – 1/3√x)^{18} is

T_{13} = T_{12 +1}

= ^{18}C_{12} (9x)^{18 - 12} (-1/3√x)^{12 }

= (-1)^{12} * 18!/(12! * 6!) * 9^{6} x^{6} (1/3)^{12 }(1/√x)^{12}

= (18 * 17 * 16 * 15 * 14 * 13 * 12!)/(12! * 6 * 5 * 4 * 3 * 2 * 1) * 3^{12} * x^{6} (1/3)^{12 }(1/x)^{6}

= 18564

Question 7:

Find the middle terms in the expansions of (3 – x^{3}/6)^{7}

Answer:

It is known that in the expansion of (a + b)^{n}, if n is odd, then there are two middle terms,

namely term {(n + 1)2}^{th} and {(n + 1)/2 + 1}^{th} term.

Therefore, the middle terms in the expansion (3 – x^{3}/6)^{7} are {(7 + 1)2}^{th} = 4^{th} and

{(7 + 1)/2 + 1}^{th} = 5^{th} term.

Now, T_{4} = T_{3 +1}

= ^{7}C_{3} (3)^{7-3} (-x^{3}/6)^{3 }

= (-1)^{3} * {7!/(3! * 4!)} * (3)^{4} * x^{9}/6^{3}

= -(7 * 6 * 5 * 4!)/(3 * 2 * 4!) * 3^{4} * 1/(2^{3} * 3^{3}) * x^{9}

= -105 x^{9}/8

T_{5} = T_{4 +1}

= ^{7}C_{4} (3)^{7-4} (-x^{3}/6)^{4 }

= (-1)^{4} * {7!/(4! * 3!)} * (3)^{3} * x^{12}/6^{3}

= (7 * 6 * 5 * 4!)/(3 * 2 * 4!) * 3^{3} * 1/(2^{4} * 3^{4}) * x^{12}

= 35x^{12}/48

Thus, the middle terms in the expansion of (3 – x^{3}/6)^{7} are: -105 x^{9}/8 and 35x^{12}/48

Question 8:

Find the middle terms in the expansions of (x/3 + 9y)^{10}

Answer:

It is known that in the expansion of (a + b)^{n}, if n is even, then the middle term is {n/2 + 1}^{th}

term.

Therefore, the middle terms in the expansion of (x/3 + 9y)^{10} is {10/2 + 1}^{th} = 6^{th} term.

Now, T_{6} = T_{5 +1}

= ^{10}C_{5} (x/3)^{10-5} (9y)^{5 }

= {10!/(5! * 5!)} * 1/3^{5} * x^{5}/3^{5} * 9^{5} * y^{5}

= (10 * 9 * 8 * 7 * 6 * 5!)/(5 * 4 * 3 * 2 * 5!) * 1/3^{5} * 3^{10} * x^{5} * y^{5}

= 252 * 3^{10} * x^{5} * y^{5}

= 61236 x^{5} y^{5}

Thus, the middle term in the expansion of (x/3 + 9y)^{10} is 61236 x^{5} y^{5}

Question 9:

In the expansion of (1 + a)^{m + n}, prove that coefficients of a^{m} and a^{n} are equal.

Answer:

^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Assuming that a^{m} occurs in the (r + 1)^{th} term of the expansion (1 + a)^{m + n}, we obtain

T_{r+1} = ^{m+n}C_{r} (1)^{m+n-r} a^{r}

Comparing the indices of a in a^{m} and in T_{r + 1}, we obtain

r = m

Therefore, the coefficient of a^{m} is

^{m+n}C_{m} = (m + n)!/{m! * (m +n - m)!} = (m + n)!/(m! * n!) ……………1

Assuming that a^{n} occurs in the (k + 1)^{th} term of the expansion (1 + a)^{m+n}, we obtain

T_{k+1} = ^{m+n}C_{k} (1)^{m+n-k} a^{k }= ^{m+n}C_{k} a^{k}

Comparing the indices of a in a^{n} and in T_{k + 1}, we obtain

k = n

Therefore, the coefficient of a^{n} is

^{m+n}C_{n} = (m + n)!/{n! * (m +n - n)!} = (m + n)!/(n! * m!) ……………2

Thus, from equation 1 and 2, it can be observed that the coefficients of a^{m} and a^{n} in the

expansion of (1 + a)^{m + n} are equal.

Question 10:

The coefficients of the (r – 1)^{th}, r^{th} and (r + 1)^{th} terms in the expansion of (x + 1)^{n} are in the ratio 1 : 3 : 5. Find n and r.

Answer:

It is known that (k + 1)^{th} term, (T_{k+1}), in the binomial expansion of (a + b)^{n} is given by

T_{k+1} = ^{n}C_{k} a^{n-k} b^{k}

Therefore, (r – 1)^{th} term in the expansion of (x + 1)^{n} is

T_{r-1} = ^{n}C_{r-2} (x)^{n-(r - 2)} 1^{r-2 }= ^{n}C_{r-2} (x)^{n - r + 2}

(r + 1) term in the expansion of (x + 1)^{n} is

T_{r+1} = ^{n}C_{r} (x)^{n-r } 1^{r }= ^{n}C_{r} (x)^{n - r }

r^{th} term in the expansion of (x + 1)^{n} is

T_{r} = ^{n}C_{r-1} (x)^{n-(r - 1)} 1^{r-1 }= ^{n}C_{r-1} (x)^{n - r + 1}

Therefore, the coefficients of the (r – 1)^{th}, r^{th}, and (r + 1)^{th} terms in the expansion of (x + 1)^{n} are

respectively. Since these coefficients are in the ratio 1:3:5, we obtain

^{n}C_{r-2} / ^{n}C_{r-1 }= 1/3 and ^{n}C_{r-1} / ^{n}C_{r }= 3/5

Now, ^{n}C_{r-2} / ^{n}C_{r-1 }= 1/3

=> [n!/{(r - 2)! * (n – r + 2)!}]/[n!/{(r - 1)! * (n – r + 1)!}] = 1/3

=> [(r - 1) * (r - 2)! * (n – r + 1)!]/ [(r - 2) * (n - r + 2)! * (n – r + 1)!] = 1/3

=> (r - 1)/(n – r + 2) = 1/3

=> 3(r - 1) = (n – r + 2)

=> 3r – 3 = n – r + 2

=> n – 4r + 5 = 0 …………1

Again, ^{n}C_{r-1} / ^{n}C_{r }= 3/5

=> [n!/{(r - 1)! * (n – r + 1)!}]/[n!/{r! * (n – r)!}] = 3/5

=> [r * (r - 1)! * (n - r)!]/ [(r - 1) * (n - r + 1)! * (n – r)!] = 3/5

=> r/(n – r + 1) = 3/5

=> 5r = 3(n – r + 1)

=> 5r = 3n – 3r + 3

=> 3n – 8r + 3 = 0 …………2

Solving equations 1 and 2, we get

Multiplying equation 1 by 3 and subtracting it from equation 2, we obtain

4r – 12 = 0

=> r = 3

Putting the value of r in equation 1, we obtain

n – 12 + 5 = 0

=> n = 7

Thus, the values of n and r are 7 and 3 respectively.

Question 11:

Prove that the coefficient of x^{n} in the expansion of (1 + x)^{2n} is twice the coefficient of x^{n} in the expansion of (1 + x)^{2n–1}

Answer:

^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Assuming that x^{n} occurs in the (r + 1)^{th} term of the expansion of (1 + x)^{2n}, we obtain

T_{r+1} = ^{2n}C_{r} (1)^{2n - r } x^{r }= ^{2n}C_{r} x^{r }

Comparing the indices of x in x^{n} and in T_{r + 1}, we obtain r = n

Therefore, the coefficient of x^{n} in the expansion of (1 + x)^{2n} is

^{2n}C_{n} = (2n)!/{n! * (2n - n)!} = (2n)!/{n! * n!} = (2n)!/(n!)^{2 } ……….1

Assuming that x^{n} occurs in the (k +1)^{th} term of the expansion (1 + x)^{2n – 1}, we obtain

T_{k+1} = ^{2n-1}C_{k} (1)^{2n-1-k } x^{k }= ^{2n-1}C_{k} (x)^{k }

Comparing the indices of x in x^{n} and T_{k + 1}, we obtain k = n

Therefore, the coefficient of x^{n} in the expansion of (1 + x)^{2n –1} is

^{2n-1}C_{n} = (2n - 1)!/{n! * (2n - 1 - n)!}

= (2n - 1)!/{n! * (n – 1)!}

= {2n * (2n - 1)!/{2n * n! * (n – 1)!}

= (2n)!/(2 * n! * n!)

= (2n)!/{2 * (n!)^{2}} ……….2

From equation 1 and 2, it is observed that

^{ 2n}C_{n} /2 = ^{2n-1}C_{n}

=> ^{2n}C_{n} = 2(^{2n-1}C_{n})

Therefore, the coefficient of x^{n} in the expansion of (1 + x)^{2n} is twice the coefficient of x^{n} in the

expansion of (1 + x)^{2n–1}.

Hence, proved.

Question 12:

Find a positive value of m for which the coefficient of x^{2} in the expansion (1 + x)^{m} is 6.

Answer:

^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Assuming that x^{2} occurs in the (r + 1)^{th} term of the expansion (1 +x)^{m}, we obtain

T_{r+1} = ^{m}C_{r} (1)^{m - r } x^{r }= ^{m}C_{r} (x)^{r }

Comparing the indices of x in x^{2} and in T_{r + 1}, we obtain r = 2

Therefore, the coefficient of x^{2} is ^{m}C_{2}

It is given that the coefficient of x^{2} in the expansion (1 + x)^{m} is 6.

So, ^{m}C_{2} = 6

=> m!/{2! * (m - 2)!} = 6

=> {m(m - 1)(m - 2)!}/{2 * (m - 2)!} = 6

=> {m(m - 1)}/2 = 6

=> m(m - 1) = 12

=> m^{2} – m – 12 = 0

=> (m - 4)(m + 3) = 0

=> m = -3, 4

Thus, the positive value of m, for which the coefficient of x^{2} in the expansion (1 + x)^{m} is 6, is 4

**Miscellaneous Exercise on Chapter 8**

Question 1:

Find a, b and n in the expansion of (a + b)^{n} if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Answer:

Given that the first three terms of the expansion are 729, 7290 and 30375, respectively.

Now T_{1} = ^{n}C_{0} * a^{n-0} * b^{0} = 729

=> a^{n} = 729 ................1

T_{2} = ^{n}C_{1} * a^{n-1} * b^{1} = 7290

=> n * a^{n-1} * b = 7290 .......2

T_{3} = ^{n}C_{2} * a^{n-2} * b^{2} = 30375

=> {n(n - 1)/2} * a^{n-2} * b^{2} = 30375 .......3

Now equation2/equation1

n * a^{n-1} * b/a^{n} = 7290/729

=> n * b/a = 10 .......4

Now equation3/equation2

{n(n - 1)/2} * a^{n-2} * b^{2} /n * a^{n-1} * b = 30375/7290

=> b(n - 1)/2a = 30375/7290

=> b(n - 1)/a = (30375*2)/7290

=> nb/a - b/a = 60750/7290

=> 10 - b/a = 6075/729 (60750 and 7290 is divided by 10)

=> 10 - b/a = 25/3 (6075 and 729 is divided by 243)

=> 10 - 25/3 = b/a

=> (30 - 25)/3 = b/a

=> 5/3 = b/a

=> b/a = 5/3 .................5

Put this value in equation 4, we get

n * 5/3 = 10

=> 5n = 30

=> n = 30/5

=> n = 6

Now put this value in equation 1, we get

a^{6} = 729

=> a^{6} = 3^{6}

=> a = 3

Now from equation 5, we get

b/3 = 5/3

=> b = 5

So, the values of a, b and n are 3, 5 and 6 respectively.

Question 2:

Find a if the coefficients of x^{2} and x^{3} in the expansion of (3 + ax)^{9} are equal.

Answer:

^{th} term, (T_{r+1}), in the binomial expansion of (a + b)^{n} is given by

T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Assuming that x^{2} occurs in the (r + 1)^{th} term in the expansion of (3 + ax)^{9}, we obtain

T_{r+1} = ^{9}C_{r} (3)^{9 - r }(ax)^{r }= ^{9}C_{r} a^{r} x^{r }

Comparing the indices of x in x^{2} and in T_{r + 1}, we obtain r = 2

Thus, the coefficient of x^{2} is

^{9}C_{2} 3^{9-2} a^{2 }= {9!/(2! * 7!)} * 3^{7} a^{2 }= 36 * 3^{7} a^{2 }

Assuming that x^{3} occurs in the (k + 1)^{th} term in the expansion of (3 + ax)^{9}, we obtain

T_{k+1} = ^{9}C_{k} (3)^{9 - r }(ax)^{k }= ^{9}C_{r} (3)^{9 - k }a^{k} x^{k}

Comparing the indices of x in x^{3} and in T_{k+ 1}, we obtain k = 3

Thus, the coefficient of x^{3} is

^{9}C_{3} 3^{9-3} a^{3 }= {9!/(3! * 6!)} * 3^{6} a^{3 }= 84 * 3^{6} a^{3 }

It is given that the coefficients of x^{2} and x^{3} are the same.

=> 84 * 3^{6} a^{3 }= 36 * 3^{7} a^{2}

=> 84a = 36 * 3

=> a = 108/84

=> a = 9/7

Thus, the required value of a is 9/7

Question 3:

Find the coefficient of x^{5} in the product (1 + 2x)^{6} (1 – x)^{7} using binomial theorem.

Answer:

Using Binomial Theorem, the expressions, (1 + 2x)^{6} and (1 – x)^{7}, can be expanded as

(1 + 2x)^{6} = ^{6}C_{0} + ^{6}C_{1} (2x) + ^{6}C_{2} (2x)^{2} + ^{6}C_{3} (2x)^{3} + ^{6}C_{4} (2x)^{4} + ^{6}C_{5} (2x)^{5} + ^{6}C_{6} (2x)^{6}

= 1 + 6(2x) + 15(2x)^{2 }+ 20(2x)^{3} + 15(2x)^{4} + 6(2x)^{5} + (2x)^{6}

= 1 + 12x + 60x^{2 }+ 160x^{3} + 240x^{4} + 192x^{5} + 64x^{6}

(1 - x)^{7} = ^{7}C_{0} - ^{7}C_{1} (x) + ^{7}C_{2} (x)^{2} - ^{7}C_{3} (x)^{3} + ^{7}C_{4} (x)^{4} - ^{7}C_{5} (x)^{5} + ^{7}C_{6} (x)^{6} - ^{7}C_{7} (x)^{7}

= 1 - 7x + 21x^{2 }- 35x^{3} + 35x^{4} - 21x^{5} + 7x^{6 }- x^{7}

Now, (1 + 2x)^{6} and (1 – x)^{7}

= (1 + 12x + 60x^{2 }+ 160x^{3} + 240x^{4} + 192x^{5} + 64x^{6}) * (1 - 7x + 21x^{2 }- 35x^{3} + 35x^{4} - 21x^{5} + 7x^{6 }- x^{7})

The complete multiplication of the two brackets is not required to be carried out. Only

those terms, which involve x^{5}, are required.

The terms containing x^{5} are

1(-21x^{5}) + (-12x)(35x^{4}) + (60x^{2})(-35x^{3}) + (160x^{3})(21x^{2}) + (240x^{4})(-7x) + (192x^{5})(1) = 171x^{5}

Thus, the coefficient of x^{5} in the given product is 171

Question 4:

If a and b are distinct integers, prove that a – b is a factor of a^{n} – b^{n}, whenever n is a positive integer. [Hint: write a^{n} = (a – b + b)^{n} and expand]

Answer:

In order to prove that (a – b) is a factor of (a^{n} – b^{n}), it has to be proved that

a^{n} – b^{n} = k (a – b), where k is some natural number

It can be written that, a = a – b + b

Now, a^{n} = (a – b + b)^{n}

= [(a – b) + b]^{n}

= ^{n}C_{0} (a - b)^{n} - ^{n}C_{1} (a - b)^{n-1} b + ^{n}C_{2} (a - b)^{n-2} b^{2} + ………+ ^{n}C_{n-1} (a - b)b^{n-1} + ^{n}C_{n} b^{n}

= (a - b)^{n} - ^{n}C_{1} (a - b)^{n-1} b + ^{n}C_{2} (a - b)^{n-2} b^{2} + ………+ ^{n}C_{n-1} (a - b)b^{n-1} + b^{n}

=> a^{n} – b^{n} = (a - b)^{n} - ^{n}C_{1} (a - b)^{n-1} b + ^{n}C_{2} (a - b)^{n-2} b^{2} + ………+ ^{n}C_{n-1} (a - b)b^{n-1}

=> a^{n} – b^{n} = (a - b)[(a - b)^{n-1} - ^{n}C_{1} (a - b)^{n-2} b + ………+ ^{n}C_{n-1} b^{n-1 }]

=> a^{n} – b^{n} = k(a - b)

Where k = [(a - b)^{n-1} - ^{n}C_{1} (a - b)^{n-2} b + ………+ ^{n}C_{n-1} b^{n-1 }] is a natural number.

This shows that (a – b) is a factor of (a^{n} – b^{n}), where n is a positive integer.

Question 5:

Evaluate. (√3 + √2)^{6} – (√3 – √2)^{6}

Answer:

Using Binomial Theorem, the expressions, (a + b)^{6} – (a – b)^{6} can be expanded as

(a + b)^{6} = ^{6}C_{0} a^{6} + ^{6}C_{1} a^{5} b + ^{6}C_{2} a^{4} b^{2} + ^{6}C_{3} a^{3} b^{3} + ^{6}C_{4} a^{2} b^{4} + ^{6}C_{5} ab^{3} + ^{6}C_{6} b^{6}

= a^{6} + 6a^{5}b + 15a^{4}b^{2} + 20a^{3}b^{3} + 15a^{2}b^{4} + 6ab^{5} + b^{6}

(a - b)^{6} = ^{6}C_{0} a^{6} - ^{6}C_{1} a^{5} b + ^{6}C_{2} a^{4} b^{2} - ^{6}C_{3} a^{3} b^{3} + ^{6}C_{4} a^{2} b^{4} - ^{6}C_{5} ab^{3} + ^{6}C_{6} b^{6}

= a^{6} - 6a^{5}b + 15a^{4}b^{2} - 20a^{3}b^{3} + 15a^{2}b^{4} - 6ab^{5} + b^{6}

Now, (a + b)^{6} – (a – b)^{6} = [a^{6} + 6a^{5}b + 15a^{4}b^{2} + 20a^{3}b^{3} + 15a^{2}b^{4} + 6ab^{5} + b^{6}] – [a^{6} - 6a^{5}b + 15a^{4}b^{2}

– 20a^{3}b^{3} + 15a^{2}b^{4} - 6ab^{5} + b^{6}]

= 2[6a^{5}b + 20a^{3}b^{3} + 6ab^{5}]

Now, put a = √3 and b = √2, we get

(√3 + √2)^{6} – (√3 – √2)^{6} = 2[6(√3)^{5}(√2) + 20(√3)^{3}(√2)^{3} + 6(√3)(√2)^{5}]

= 2[54√6 + 120√6 + 24√6]

= 2 * 198√6

= 396√6

Question 6:

Find the value of {a^{2} + √(a^{2} - 1)}^{4} + {a^{2} + √(a^{2} - 1)}^{4}

Answer 6:

Firstly, the expression (x + y)^{4} + (x – y)^{4} is simplified by using Binomial Theorem.

(x + y)^{4} = ^{4}C_{0} x^{4} + ^{4}C_{1} x^{3} y + ^{4}C_{2} x^{2} y^{2} + ^{4}C_{3} xy^{3} + ^{4}C_{4} y^{4}

= x^{4} + 4x^{3 }y + 6x^{2}y^{2} + 4xy^{3} + y^{4}

(x - y)^{4} = ^{4}C_{0} x^{4} + ^{4}C_{1} x^{3} y + ^{4}C_{2} x^{2} y^{2} + ^{4}C_{3} xy^{3} + ^{4}C_{4} y^{4}

= x^{4} - 4x^{3 }y + 6x^{2}y^{2} - 4xy^{3} + y^{4}

Now, (x + y)^{4} + (x – y)^{4} = [x^{4} + 4x^{3 }y + 6x^{2}y^{2} + 4xy^{3} + y^{4}] – [x^{4} - 4x^{3 }y + 6x^{2}y^{2} - 4xy^{3} + y^{4}]

= 2[x^{4} + 6x^{2}y^{2} + y^{4}]

Put x = a^{2} and y = √(a^{2} - 1)}, we get

a^{2} + √(a^{2} - 1)}^{4} + {a^{2} + √(a^{2} - 1)}^{4} = 2[(a^{2})^{4} + 6(a^{2})^{2}{(a^{2} - 1)}^{2} + {(a^{2} - 1)}^{4}]

= 2[a^{8} + 6a^{4} (a^{2} - 1) + (a^{2} - 1)^{2}]

= 2[a^{8} + 6a^{6} - 6a^{4} + a^{4} + 1 – 2a^{2}]

= 2[a^{8} + 6a^{6} - 5a^{4} - 2a^{2} + 1]

= 2a^{8} + 12a^{6} - 10a^{4} - 4a^{2} + 2

Question 7:

Find an approximation of (0.99)^{5} using the first three terms of its expansion.

Answer:

0.99 can be written as,

0.99 = 1 – 0.01

Now, (0.99)^{5} = (1 – 0.01)^{5}

= ^{5}C_{0} (1)^{5} - ^{5}C_{1} (1)^{4} (0.01) + ^{5}C_{2} (1)^{3}(0.01)^{2}

= 1 – 5(0.01) + 10(0.01)^{2}

= 1 – 0.05 + 0.001

= 0.951

Thus, the approximation of (0.99)^{5} using the first three terms of its expansion is 0.951

Question 8:

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (^{4}√2 + 1/^{4}√3)^{n} is √6 : 1

Answer 8:

In the expansion, (a + b)^{n} = ^{n}C_{0} a^{n} + ^{n}C_{1} a^{n-1} b + ^{n}C_{2} a^{n-2} b^{2} + …………….. + ^{n}C_{n} b^{n}

Fifth term from the beginning = ^{n}C_{4} a^{n-4} b^{4}

Fifth term from the end = ^{n}C_{n-4} a^{4} b^{n-4}

Therefore, it is evident that in the expansion of (^{4}√2 + 1/^{4}√3)^{n} the fifth term from the beginning

is ^{n}C_{4} (^{4}√2)^{n-4} (1/^{4}√3)^{4 }and the fifth term from the end is ^{n}C_{n-4} (^{4}√2)^{4} (1/^{4}√3)^{n-4}

Now, ^{n}C_{4} (^{4}√2)^{n-4} (1/^{4}√3)^{4 }= ^{n}C_{4} {(^{4}√2)^{n}/(^{4}√2)^{4}}(1/3)

= ^{n}C_{4} {(^{4}√2)^{n}/2}(1/3)

= [n!/{(6 * 4! * (n - 4)!}] * (^{4}√2)^{n} ……………1

and ^{n}C_{n-4} (^{4}√2)^{4} (1/^{4}√3)^{n-4 }= ^{n}C_{n-4} {2 * (^{4}√3)^{4}/(^{4}√3)^{n}}

= ^{n}C_{n-4} {2 * 3/(^{4}√3)^{n} }

= [6n!/{(4! * (n - 4)!}] * {1/(^{4}√3)^{n}} ……………2

It is given that the ratio of the fifth term from the beginning to the fifth term from the end is

√6 : 1. Therefore, from equation 1 and 2, we get

[n!/{(6 * 4! * (n - 4)!}] * (^{4}√2)^{n} : [6n!/{(4! * (n - 4)!}] * {1/(^{4}√3)^{n}} = √6 : 1

=> [(^{4}√2)^{n} /6] * [(^{4}√3)^{n}/6] = √6

=> (^{4}√6)^{n} /36 = √6

=> (^{4}√6)^{n} = 36√6

=> 6^{n/4} = 6^{5/2}

=> n/4 = 5/2

=> n/2 = 5

=> n = 10

Thus, the value of n is 10

Question 9:

Expand using Binomial Theorem (1 + x/2 – 2/x)^{4}, x ≠ 0

Answer:

Using Binomial Theorem, the given expression (1 + x/2 – 2/x)^{4} can be expanded as

(1 + x/2 – 2/x)^{4} = {(1 + x/2) – 2/x}^{4}

= ^{4}C_{0} (1 + x/2)^{4} - ^{4}C_{1} (1 + x/2)^{3} (2/x) + ^{4}C_{2} (1 + x/2)^{2} (2/x)^{2} - ^{4}C_{3} (1 + x/2)( 2/x)^{3} + ^{4}C_{4} (2/x)^{4}

= (1 + x/2)^{4} - 4(1 + x/2)^{3} (2/x) + 6(1 + x/2)^{2} (4/x^{2}) - 4(1 + x/2)(8/x^{3}) + 16/x^{4}

= (1 + x/2)^{4} - (1 + x/2)^{3} (8/x) + (1 + x + x^{2}/4) (24/x^{2}) - (1 + x/2)(32/x^{3}) + 16/x^{4}

= (1 + x/2)^{4} - (1 + x/2)^{3} (8/x) + 24/x^{2} + 24/x + 6 – 32/x^{3} - 16/x^{2} + 16/x^{4}

= (1 + x/2)^{4} - (1 + x/2)^{3} (8/x) + 8/x^{2} + 24/x + 6 – 32/x^{3} + 16/x^{4}

=> (1 + x/2 – 2/x)^{4} = (1 + x/2)^{4} - (1 + x/2)^{3} (8/x) + 8/x^{2} + 24/x + 6 – 32/x^{3} + 16/x^{4} ……….1

Again by using Binomial Theorem, we obtain

(1 + x/2)^{4} = ^{4}C_{0} (1)^{4} + ^{4}C_{1} (1)^{3} (x/2) + ^{4}C_{2} (1)^{2} (x/2)^{2} + ^{4}C_{3} (1)( x/2)^{3} + ^{4}C_{4} (x/2)^{4}

= 1 + 4 * (x/2) + 6 * (x^{2}/4) + 4 * (x^{3}/8) + x^{4}/16

=> (1 + x/2)^{4} = 1 + 2x + 3x^{2}/2 + x^{3}/2 + x^{4}/16 …………..2

(1 + x/2)^{3} = ^{3}C_{0} (1)^{3} + ^{3}C_{1} (1)^{2} (x/2) + ^{3}C_{2} (1)(x/2)^{2} + ^{3}C_{3} (x/2)^{3}

=> (1 + x/2)^{3} = 1 + 3x/2 + 3x^{2}/4 + x^{3}/8 …………..3

From equation 1, 2, and 3, we get

(1 + x/2 – 2/x)^{4} = 1 + 4 * (x/2) + 6 * (x^{2}/4) + 4 * (x^{3}/8) + x^{4}/16 – [1 + 3x/2 + 3x^{2}/4 + x^{3}/8](8/x) +

8/x^{2} + 24/x + 6 - 32/x^{3} + 16/x^{4}

= 1 + 2x + 3x^{2}/2 + x^{3}/2 + x^{4}/16 – 8/x - 12 – 6x – x^{2} + 8/x^{2} + 24/x + 6 - 32/x^{3} + 16/x^{4}

= 16/x + 8/x^{2} – 32/x^{3} +16/x^{4} + x^{2}/2 + x^{3}/2 + x^{4}/16 - 5

Question 10:

Find the expansion of (3x^{2} – 2ax + 3a^{2})^{3} using binomial theorem.

Answer:

Using Binomial Theorem, the given expression (3x^{2} – 2ax + 3a^{2})^{3} can be expanded as

(3x^{2} – 2ax + 3a^{2})^{3} = [(3x^{2} – 2ax) + 3a^{2}]^{3}

= ^{3}C_{0} (3x^{2} – 2ax)^{3} + ^{3}C_{1} (3x^{2} – 2ax)^{2} (3a^{2}) + ^{3}C_{2} (3x^{2} – 2ax) (3a^{2})^{2} + ^{3}C_{3} (3a^{2})^{3}

= (3x^{2} – 2ax)^{3} + 3(9x^{2} – 12ax^{3} + 4a^{2} x^{2})(3a^{2}) + 3(3x^{2} – 2ax) (9a^{4}) + 27a^{6}

= (3x^{2} – 2ax)^{3} + 81a^{2} x^{4} – 108a^{3} x^{3} + 36a^{4} x^{2} + 81a^{4} x^{2} – 54a^{5} x + 27a^{6}

=> (3x^{2} – 2ax + 3a^{2})^{3} = (3x^{2} – 2ax)^{3} + 81a^{2} x^{4} – 108a^{3} x^{3} + 117a^{4} x^{2} – 54a^{5} x + 27a^{6 } ………….1

Again by using Binomial Theorem, we obtain

(3x^{2} – 2ax)^{3} = ^{3}C_{0} (3x^{2})^{3} - ^{3}C_{1} (3x^{2})^{2} (2ax) + ^{3}C_{2} (3x^{2})( 2ax)^{2} - ^{3}C_{3} (2ax)^{3}

=> (3x^{2} – 2ax)^{3} = 27x^{6} - 3(9x^{4})(2ax) + 3(3x^{2})(4a^{2} x^{2}) - 8a^{3} x^{3}

=> (3x^{2} – 2ax)^{3} = 27x^{6} – 54ax^{5} + 36a^{2} x^{4} - 8a^{3} x^{3 } …………..2

From equation 1 and 2, we get

(3x^{2} – 2ax + 3a^{2})^{3} = 27x^{6} – 54ax^{5} + 36a^{2} x^{4} - 8a^{3} x^{3} + 81a^{2} x^{4} – 108a^{3} x^{3} + 117a^{4} x^{2} – 54a^{5} x + 27a^{6 }

= 27x^{6} – 54ax^{5} + 117a^{2} x^{4} - 116a^{3} x^{3} + 117a^{4} x^{2} – 54a^{5} x + 27a^{6}

.