Class 11 - Maths - Binomial Theorem

Exercise 8.1

Question 1:

Expand the expression (1– 2x)5

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

(1– 2x)5 = 5C0 (1)5 - 5C1 (1)4 (2x) + 5C2 (1)3 (2x)2 - 5C4 (1)1 (2x)4 + 5C5 (2x)5

= 1 – 5(2x) + 10(4x2) – 10(8x3) + 5(16x4) – 32x5

= 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5

Question 2:

Expand the expression (2/x – x/2)5

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

(2/x– x/2)5 = 5C0 (2/x)5 - 5C1 (2/x)4 (x/2) + 5C2 (2/x)3 (x/2)2 - 5C4 (2/x)1 (x/2)4 + 5C5 (x/2)5

= 32/x5 – 5(16/x4)(x/2) + 10(8/x3)(x2/4) – 10(4/x2)(x3/8) + 5(2/x)(x4/16) – x5/32

= 32/x5 – 40/x3 + 20/x – 5x + 5x3/8 - x5/32

Question 3:

Expand the expression (2x – 3)6

By using Binomial Theorem, the expression can be expanded as

(2x – 3)6 = 6C0 (2x)6 - 6C1 (2x)5 (3) + 6C2 (2x)4 (3)2 - 6C3 (2x)3 (3)2 + 6C4 (2x)2 (3)4 - 6C5 (2x)1 (3)5 +

6C6 (3)6

= 64x6 – 6(32x5)* 3 + 15(16x4) * 9 – 20(8x3) * 27 + 15(4x2) * 81 – 6 * 2x * 243 + 729

= 64x6 – 576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729

Question 4:

Expand the expression (x/3 + 1/x)5

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

(x/3 + 1/x)5 = 5C0 (x/3)5 + 5C1 (x/3)4 (1/x) + 5C2 (x/3)3 (1/x)2 + 5C4 (x/3)1 (1/x)4 + 5C5 (1/x)5

= x5/343 + 5(x4/81)(1/x) + 10(x3/27)(1/x2) + 10(x2/9)(1/x3) + 5(x/3)(1/x4) + 1/x5

= x5/343 + 5x3/81 + 10x/27 +10/9x + 5/3x3 + 1/x5

Question 5:

Expand the expression (2/x – x/2)5

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

(x + 1/x)6 = 6C0 (x)6 + 6C1 (x)5 (1/x) + 6C2 (x)4 (1/x)2 + 6C3 (x)3 (1/x)4 + 6C4 (x)2 (1/x)4 + 6C5 (x) (1/x)5

+ 6C6 (1/x)6

= x6 + 6(x)5(1/x) + 14(x)4(1/x2) + 20(x)3(1/x3) + 15(x)2(1/x4) + 6(x)(1/x5) + 1/x6

= x6 + 6x4 + 15x2 + 20 + 15/x2 + 6/x4 + 1/x6

Question 6:

Using Binomial Theorem, evaluate (96)3

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

= 3C0 (100)3 - 3C1 (100)2 (4) + 3C2 (100)(4)2 - 3C3 (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

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

= 5C0 (100)5 + 5C1 (100)4 (2) + 5C2 (100)3(2)2 + 5C3 (100)2 (2)3 + 5C4 (100)(2)4 + 5C5 (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

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

= 4C0 (100)4 + 4C1 (100)3 (1) + 4C2 (100)2(1)2 + 4C3 (100) (1)3 + 4C4 (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

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

= 5C0 (100)5 - 5C1 (100)4 (1) + 5C2 (100)3(1)2 - 5C3 (100)2 (1)3 + 5C4 (100)(1)4 - 5C5 (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.

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

= 10000C0 - 10000C1 (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

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

(a + b)4 = 4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4

(a - b)4 = 4C0 a4 - 4C1 a3 b + 4C2 a2 b2 - 4C3 a b3 + 4C4 b4

Now, (a + b)4 – (a – b)4 = [4C0 a4 + 4C1 a3 b + 4C2 a2 b2 + 4C3 a b3 + 4C4 b4] – [4C0 a4 - 4C1 a3 b + 4C2 a2

b2 - 4C3 a b3 + 4C4 b4]

= 2[4C1 a3 b + 4C3 a b3]

= 8ab(a2 + b2)

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

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

(x + 1)6 = 6C0 x6 + 6C1 x5 + 6C2 x4 + 6C3 x3 + 6C4 x2 + 6C5 x + 6C6

(x - 1)6 = 6C0 x6 - 6C1 x5 + 6C2 x4 - 6C3 x3 + 6C4 x2 - 6C5 x + 6C6

Now, (x + 1)6 + (x – 1)6 = 2[6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6]

= 2[x6 + 15x4 + 15x2 + 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 9n+1 - 8n - 9 is divisible by 64, whenever n is a positive integer.

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

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

By Binomial Theorem,

(1 + a)m = mC0 + mC1 a + mC2 a2 +………..+ mCm am

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

(1 + 8)n+1 = n+1C0 + n+1C1 8 + n+1C2 82 +………..+ n+1Cn+1 8n+1

=> 9n+1 = 1 + 8(n + 1) + 82 [n+1C2 + n+1C3 8 +………..+ n+1Cn+1 8n-1]

=> 9n+1 = 1 + 8n + 8 + 64[n+1C2 + n+1C3 8 +………..+ n+1Cn+1 8n-1]

=> 9n+1 = 9 + 8n + 64[n+1C2 + n+1C3 8 +………..+ n+1Cn+1 8n-1]

=> 9n+1 - 8n – 9 = 64[n+1C2 + n+1C3 8 +………..+ n+1Cn+1 8n-1]

=> 9n+1 - 8n – 9 = 64k, where k = [n+1C2 + n+1C3 8 +………..+ n+1Cn+1 8n-1], is a natural number.

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

Question 14:

Prove that: r nCr = 4n

By Binomial Theorem,

nCr an-r br = (a + b)n

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

nCr 1n-r 3r = (1 + 3)n

=> r nCr = 4n

Hence, proved.

Exercise 8.2

Question 1:

Find the coefficient of x5 in (x + 3)8

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

Tr+1 = nCr an-r br

Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, we obtain

T8+1 = 8Cr a8 - r 3r

Comparing the indices of x in x5 and in Tr +1,we obtain r = 3

Thus, the coefficient of x5 is

8C3 33 = 8!/(3! * 5!) * 33 = (8 * 7 * 6 * 5!)/(3 * 2 * 1 * 5!) * 33

= 8 * 7 * 27

= 1512

Question 2:

Find the coefficient of a5 b7 in (a – 2b)12

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

Tr+1 = nCr an-r br

Assuming that a5 b7 occurs in the (r + 1)th term of the expansion (a – 2b)12, we obtain

T8+1 = 12Cr a12 - r (-2b)r = 12Cr (-2)r a12 - r (b)r

Comparing the indices of x in a5 b7 and in Tr +1, we obtain r = 7

Thus, the coefficient of a5 b7 is

12C7 (-2)7 = -12!/(7! * 5!) * 27 = -(12 * 11 * 10 * 9 * 8 * 7!)/(5 * 4 * 3 * 2 * 1 * 5!) * 27

= -792 * 128

= -101376

Question 3:

Write the general term in the expansion of (x2 – y)6

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 Tr+1 = nCr an-r br.

Thus, the general term in the expansion of (x2 – y)6 is

Tr +1 = 6Cr (x2)6 - r (-y)r = (-1)r 6Cr c12 -2 r yr

Question 4:

Write the general term in the expansion of (x2 – yx)12, x ≠ 0

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 Tr+1 = nCr an-r br.

Thus, the general term in the expansion of (x2 – yx)12 is

Tr +1 = 12Cr (x2)12 - r (-yx)r = (-1)r 12Cr x24 -2r yr xr = (-1)r 12Cr x24 - r yr

Question 5:

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

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

Tr+1 = nCr an-r br

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

T4 = T3 +1 = 12C3 (x2)12 - 3 (-2y)3 = (-1)3 * 12!/(9! * 3!) x9 23 y3

= -(12 * 11 * 10)/(3 * 2)  * 23 * x9 y3

= -1760 x9 y3

Question 6:

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

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

Tr+1 = nCr an-r br

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

T13 = T12 +1

= 18C12 (9x)18 - 12 (-1/3√x)12

= (-1)12 * 18!/(12! * 6!) * 96 x6 (1/3)12 (1/√x)12

= (18 * 17 * 16 * 15 * 14 * 13 * 12!)/(12! * 6 * 5 * 4 * 3 * 2 * 1) * 312 * x6 (1/3)12 (1/x)6

= 18564

Question 7:

Find the middle terms in the expansions of (3 – x3/6)7

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 – x3/6)7 are {(7 + 1)2}th = 4th and

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

Now, T4 = T3 +1

= 7C3 (3)7-3 (-x3/6)3

= (-1)3 * {7!/(3! * 4!)} * (3)4  * x9/63

= -(7 * 6 * 5 * 4!)/(3 * 2 * 4!) * 34 * 1/(23 * 33) * x9

= -105 x9/8

T5 = T4 +1

= 7C4 (3)7-4 (-x3/6)4

= (-1)4 * {7!/(4! * 3!)} * (3)3 * x12/63

= (7 * 6 * 5 * 4!)/(3 * 2 * 4!) * 33 * 1/(24 * 34) * x12

= 35x12/48

Thus, the middle terms in the expansion of (3 – x3/6)7 are: -105 x9/8 and 35x12/48

Question 8:

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

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 = 6th term.

Now, T6 = T5 +1

= 10C5 (x/3)10-5 (9y)5

= {10!/(5! * 5!)} * 1/35 * x5/35 * 95 * y5

= (10 * 9 * 8 * 7 * 6 * 5!)/(5 * 4 * 3 * 2 * 5!) * 1/35 * 310 * x5 * y5

= 252 * 310 * x5 * y5

= 61236 x5 y5

Thus, the middle term in the expansion of (x/3 + 9y)10 is 61236 x5 y5

Question 9:

In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.

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

Tr+1 = nCr an-r br

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

Tr+1 = m+nCr (1)m+n-r ar

Comparing the indices of a in am and in Tr + 1, we obtain

r = m

Therefore, the coefficient of am is

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

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

Tk+1 = m+nCk (1)m+n-k ak = m+nCk ak

Comparing the indices of a in an and in Tk + 1, we obtain

k = n

Therefore, the coefficient of an is

m+nCn = (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 am and an in the

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

Question 10:

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

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

Tk+1 = nCk an-k bk

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

Tr-1 = nCr-2 (x)n-(r - 2) 1r-2 = nCr-2 (x)n - r + 2

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

Tr+1 = nCr (x)n-r  1r = nCr (x)n - r

rth term in the expansion of (x + 1)n is

Tr = nCr-1 (x)n-(r - 1) 1r-1 = nCr-1 (x)n - r + 1

Therefore, the coefficients of the (r – 1)th, rth, 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

nCr-2 / nCr-1 = 1/3 and nCr-1 / nCr = 3/5

Now, nCr-2 / nCr-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, nCr-1 / nCr = 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 xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1

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

Tr+1 = nCr an-r br

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

Tr+1 = 2nCr (1)2n - r  xr = 2nCr xr

Comparing the indices of x in xn and in Tr + 1, we obtain r = n

Therefore, the coefficient of xn in the expansion of (1 + x)2n is

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

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

Tk+1 = 2n-1Ck (1)2n-1-k  xk = 2n-1Ck (x)k

Comparing the indices of x in xn and Tk + 1, we obtain k = n

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

2n-1Cn = (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

2nCn /2 = 2n-1Cn

=> 2nCn = 2(2n-1Cn)

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

expansion of (1 + x)2n–1.

Hence, proved.

Question 12:

Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.

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

Tr+1 = nCr an-r br

Assuming that x2 occurs in the (r + 1)th term of the expansion (1 +x)m, we obtain

Tr+1 = mCr (1)m - r  xr = mCr (x)r

Comparing the indices of x in x2 and in Tr + 1, we obtain r = 2

Therefore, the coefficient of x2 is mC2

It is given that the coefficient of x2 in the expansion (1 + x)m is 6.

So, mC2 = 6

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

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

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

=> m(m - 1) = 12

=> m2 – m – 12 = 0

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

=> m = -3, 4

Thus, the positive value of m, for which the coefficient of x2 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.

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

Now T1 = nC0 * an-0 * b0 = 729

=> an = 729  ................1

T2 = nC1 * an-1 * b1 = 7290

=> n * an-1 * b = 7290 .......2

T3 = nC2 * an-2 * b2 = 30375

=> {n(n - 1)/2} * an-2 * b2 = 30375 .......3

Now equation2/equation1

n * an-1 * b/an  = 7290/729

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

Now equation3/equation2

{n(n - 1)/2} * an-2 * b2 /n * an-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

a6 = 729

=> a6 = 36

=> 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 x2 and x3 in the expansion of (3 + ax)9 are equal.

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

Tr+1 = nCr an-r br

Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9, we obtain

Tr+1 = 9Cr (3)9 - r (ax)r = 9Cr ar xr

Comparing the indices of x in x2 and in Tr + 1, we obtain r = 2

Thus, the coefficient of x2 is

9C2 39-2 a2 = {9!/(2! * 7!)} * 37 a2 = 36 * 37 a2

Assuming that x3 occurs in the (k + 1)th term in the expansion of (3 + ax)9, we obtain

Tk+1 = 9Ck (3)9 - r (ax)k = 9Cr (3)9 - k ak xk

Comparing the indices of x in x3 and in Tk+ 1, we obtain k = 3

Thus, the coefficient of x3 is

9C3 39-3 a3 = {9!/(3! * 6!)} * 36 a3 = 84 * 36 a3

It is given that the coefficients of x2 and x3 are the same.

=> 84 * 36 a3 = 36 * 37 a2

=> 84a = 36 * 3

=> a = 108/84

=> a = 9/7

Thus, the required value of a is 9/7

Question 3:

Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.

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

(1 + 2x)6 = 6C0 + 6C1 (2x) + 6C2 (2x)2 + 6C3 (2x)3 + 6C4 (2x)4 + 6C5 (2x)5 + 6C6 (2x)6

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

= 1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6

(1 - x)7 = 7C0 - 7C1 (x) + 7C2 (x)2 - 7C3 (x)3 + 7C4 (x)4 - 7C5 (x)5 + 7C6 (x)6 - 7C7 (x)7

= 1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7

Now, (1 + 2x)6 and (1 – x)7

= (1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6) * (1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7)

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

those terms, which involve x5, are required.

The terms containing x5 are

1(-21x5) + (-12x)(35x4) + (60x2)(-35x3) + (160x3)(21x2) + (240x4)(-7x) + (192x5)(1) = 171x5

Thus, the coefficient of x5 in the given product is 171

Question 4:

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

In order to prove that (a – b) is a factor of (an – bn), it has to be proved that

an – bn = k (a – b), where k is some natural number

It can be written that, a = a – b + b

Now, an = (a – b + b)n

= [(a – b) + b]n

= nC0 (a - b)n - nC1 (a - b)n-1 b + nC2 (a - b)n-2 b2 + ………+ nCn-1 (a - b)bn-1 + nCn bn

= (a - b)n - nC1 (a - b)n-1 b + nC2 (a - b)n-2 b2 + ………+ nCn-1 (a - b)bn-1 + bn

=> an – bn = (a - b)n - nC1 (a - b)n-1 b + nC2 (a - b)n-2 b2 + ………+ nCn-1 (a - b)bn-1

=> an – bn = (a - b)[(a - b)n-1 - nC1 (a - b)n-2 b + ………+ nCn-1 bn-1 ]

=> an – bn = k(a - b)

Where k = [(a - b)n-1 - nC1 (a - b)n-2 b + ………+ nCn-1 bn-1 ] is a natural number.

This shows that (a – b) is a factor of (an – bn), where n is a positive integer.

Question 5:

Evaluate. (√3 + √2)6 – (√3 – √2)6

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

(a + b)6 = 6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 ab3 + 6C6 b6

= a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6

(a - b)6 = 6C0 a6 - 6C1 a5 b + 6C2 a4 b2 - 6C3 a3 b3 + 6C4 a2 b4 - 6C5 ab3 + 6C6 b6

= a6 - 6a5b + 15a4b2 - 20a3b3 + 15a2b4 - 6ab5 + b6

Now, (a + b)6 – (a – b)6 = [a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6] – [a6 - 6a5b + 15a4b2

– 20a3b3 + 15a2b4 - 6ab5 + b6]

= 2[6a5b + 20a3b3 + 6ab5]

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 {a2 + √(a2 - 1)}4 + {a2 + √(a2 - 1)}4

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

(x + y)4 = 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 xy3 + 4C4 y4

= x4 + 4x3 y + 6x2y2 + 4xy3 + y4

(x - y)4 = 4C0 x4 + 4C1 x3 y + 4C2 x2 y2 + 4C3 xy3 + 4C4 y4

= x4 - 4x3 y + 6x2y2 - 4xy3 + y4

Now, (x + y)4 + (x – y)4 = [x4 + 4x3 y + 6x2y2 + 4xy3 + y4] – [x4 - 4x3 y + 6x2y2 - 4xy3 + y4]

= 2[x4 + 6x2y2 + y4]

Put x = a2 and y = √(a2 - 1)}, we get

a2 + √(a2 - 1)}4 + {a2 + √(a2 - 1)}4 = 2[(a2)4 + 6(a2)2{(a2 - 1)}2 + {(a2 - 1)}4]

= 2[a8 + 6a4 (a2 - 1) + (a2 - 1)2]

= 2[a8 + 6a6 - 6a4 + a4 + 1 – 2a2]

= 2[a8 + 6a6 - 5a4 - 2a2 + 1]

= 2a8 + 12a6 - 10a4 - 4a2 + 2

Question 7:

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

0.99 can be written as,

0.99 = 1 – 0.01

Now, (0.99)5 = (1 – 0.01)5

= 5C0 (1)5 - 5C1 (1)4 (0.01) + 5C2 (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

In the expansion, (a + b)n = nC0 an + nC1 an-1 b + nC2 an-2 b2 + …………….. + nCn bn

Fifth term from the beginning = nC4 an-4 b4

Fifth term from the end = nCn-4 a4 bn-4

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

is nC4 (4√2)n-4 (1/4√3)4 and the fifth term from the end is nCn-4 (4√2)4 (1/4√3)n-4

Now, nC4 (4√2)n-4 (1/4√3)4 = nC4 {(4√2)n/(4√2)4}(1/3)

= nC4 {(4√2)n/2}(1/3)

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

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

= nCn-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

=> 6n/4 = 65/2

=> 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

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

= 4C0 (1 + x/2)4 - 4C1 (1 + x/2)3 (2/x) + 4C2 (1 + x/2)2 (2/x)2 - 4C3 (1 + x/2)( 2/x)3 + 4C4 (2/x)4

= (1 + x/2)4 - 4(1 + x/2)3 (2/x) + 6(1 + x/2)2 (4/x2) - 4(1 + x/2)(8/x3) + 16/x4

= (1 + x/2)4 - (1 + x/2)3 (8/x) + (1 + x + x2/4) (24/x2) - (1 + x/2)(32/x3) + 16/x4

= (1 + x/2)4 - (1 + x/2)3 (8/x) + 24/x2 + 24/x + 6 – 32/x3 - 16/x2 + 16/x4

= (1 + x/2)4 - (1 + x/2)3 (8/x) + 8/x2 + 24/x + 6 – 32/x3 + 16/x4

=> (1 + x/2 – 2/x)4 = (1 + x/2)4 - (1 + x/2)3 (8/x) + 8/x2 + 24/x + 6 – 32/x3 + 16/x4  ……….1

Again by using Binomial Theorem, we obtain

(1 + x/2)4 = 4C0 (1)4 + 4C1 (1)3 (x/2) + 4C2 (1)2 (x/2)2 + 4C3 (1)( x/2)3 + 4C4 (x/2)4

= 1 + 4 * (x/2) + 6 * (x2/4) + 4 * (x3/8) + x4/16

=> (1 + x/2)4 = 1 + 2x + 3x2/2 + x3/2 + x4/16  …………..2

(1 + x/2)3 = 3C0 (1)3 + 3C1 (1)2 (x/2) + 3C2 (1)(x/2)2 + 3C3 (x/2)3

=> (1 + x/2)3 = 1 + 3x/2 + 3x2/4 + x3/8  …………..3

From equation 1, 2, and 3, we get

(1 + x/2 – 2/x)4 = 1 + 4 * (x/2) + 6 * (x2/4) + 4 * (x3/8) + x4/16 – [1 + 3x/2 + 3x2/4 + x3/8](8/x) +

8/x2 + 24/x + 6 - 32/x3 + 16/x4

= 1 + 2x + 3x2/2 + x3/2 + x4/16 – 8/x - 12 – 6x – x2 + 8/x2 + 24/x + 6 - 32/x3 + 16/x4

= 16/x + 8/x2 – 32/x3 +16/x4 + x2/2 + x3/2 + x4/16 - 5

Question 10:

Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.

Using Binomial Theorem, the given expression (3x2 – 2ax + 3a2)3 can be expanded as

(3x2 – 2ax + 3a2)3 = [(3x2 – 2ax) + 3a2]3

= 3C0 (3x2 – 2ax)3 + 3C1 (3x2 – 2ax)2 (3a2) + 3C2 (3x2 – 2ax) (3a2)2 + 3C3 (3a2)3

= (3x2 – 2ax)3 + 3(9x2 – 12ax3 + 4a2 x2)(3a2) + 3(3x2 – 2ax) (9a4) + 27a6

= (3x2 – 2ax)3 + 81a2 x4 – 108a3 x3 + 36a4 x2 + 81a4 x2 – 54a5 x + 27a6

=> (3x2 – 2ax + 3a2)3 = (3x2 – 2ax)3 + 81a2 x4 – 108a3 x3 + 117a4 x2 – 54a5 x + 27a6  ………….1

Again by using Binomial Theorem, we obtain

(3x2 – 2ax)3 = 3C0 (3x2)3 - 3C1 (3x2)2 (2ax) + 3C2 (3x2)( 2ax)2 - 3C3 (2ax)3

=> (3x2 – 2ax)3 = 27x6 - 3(9x4)(2ax) + 3(3x2)(4a2 x2) - 8a3 x3

=> (3x2 – 2ax)3 = 27x6 – 54ax5 + 36a2 x4 - 8a3 x3   …………..2

From equation 1 and 2, we get

(3x2 – 2ax + 3a2)3 = 27x6 – 54ax5 + 36a2 x4 - 8a3 x3 + 81a2 x4 – 108a3 x3 + 117a4 x2 – 54a5 x + 27a6

= 27x6 – 54ax5 + 117a2 x4 - 116a3 x3 + 117a4 x2 – 54a5 x + 27a6