Class 11 - Maths - Mathematical Induction

** Exercise 4.1**

Question 1:

Prove the following by using the principle of mathematical induction for all n є N:

1 + 3 + 3^{2} + 3^{3} + …… + 3^{n-1} = (3^{n} – 1)/2

Answer 1:

Let the given statement be P(n), i.e.,

P(n): 1 + 3 + 3^{2} + 3^{3} + …… + 3^{n-1} = (3^{n} – 1)/2

For n = 1, we have

P(1):= (3^{1} – 1)/2 = (3 - 1)/2 = 2/2 = 1, which is true.

Let P(k) be true for some positive integer k, i.e.,

1 + 3 + 3^{2} + 3^{3} + …… + 3^{k-1} = (3^{k} – 1)/2…………..1

We shall now prove that P(k + 1) is true.

Consider

1 + 3 + 3^{2} + ... + 3^{k–1} + 3^{(k+1) – 1} = (1 + 3 + 3^{2} +... + 3^{k–1}) + 3^{k}

= (3^{k} – 1)/2 + 3^{k}

= {(3^{k} – 1) + 2 * 3^{k}}/2

= {(1 + 2)3^{k} – 1}/2

= {3 * 3^{k} – 1}/2

= (3^{k+1} – 1)/2

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 2:

Prove the following by using the principle of mathematical induction for all n є N:

1^{3} + 2^{3} + 3^{3} + ………..+ n^{3} = {n(n + 1)/2}^{2}

Answer:

Let the given statement be P(n), i.e.,

P(n): 1^{3} + 2^{3} + 3^{3} + ………..+ n^{3} = {n(n + 1)/2}^{2}

For n = 1, we have

P(1): 1^{3} =1= {1(1 + 1)/2}^{2} = {(1 * 2)/2}^{2} = 1^{2} = 1, which is true.

Let P(k) be true for some positive integer k, i.e.,

1^{3} + 2^{3} + 3^{3} + ………..+ k^{3} = {k(k + 1)/2}^{2} ……………..1

We shall now prove that P(k + 1) is true.

Consider

1^{3} + 2^{3} + 3^{3} + ... + k^{3} + (k + 1)^{3}

= (1^{3} + 2^{3} + 3^{3} + .... + k^{3}) + (k + 1)^{3}

= {k(k + 1)/2}^{2} + (k + 1)^{3}

= k^{2}(k + 1)^{2}/4 + (k + 1)^{3}

= {k^{2}(k + 1)^{2} + 4(k + 1)^{3 }}/4

= (k + 1)^{2}{k^{2} + 4(k + 1)}/4

= (k + 1)^{2}{k^{2} + 4k + 4}/4

= {(k + 1)^{2 }(k + 2)^{2}}/4

= {(k + 1)^{2 }(k + 1 + 1)^{2}}/4

= {(k + 1)(k + 1 + 1)/2}^{2}

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 3:

Prove the following by using the principle of mathematical induction for all n є N:

1 + 1/(1 + 2) + 1/(1 + 2 + 3) + …………..+ 1/(1 + 2 + 3 + ……..+ n) = 2n/(n + 1)

Answer:

Let the given statement be P(n), i.e.,

P(n): 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + …………..+ 1/(1 + 2 + 3 + ……..+ n) = 2n/(n + 1)

For n = 1, we have

P(1): 1 = (2 * 1)/(1 + 1) = 2/2 = 1, which is true.

Let P(k) be true for some positive integer k, i.e.,

1 + 1/(1 + 2) + 1/(1 + 2 + 3) + …………..+ 1/(1 + 2 + 3 + ……..+ k) = 2k/(k + 1)

We shall now prove that P(k + 1) is true.

Consider

1 + 1/(1 + 2) + 1/(1 + 2 + 3) + …………..+ 1/{1 + 2 + 3 + ……..+ k + (k + 1)}

= {1 + 1/(1 + 2) + 1/(1 + 2 + 3) + ……+ 1/(1 + 2 + 3 + ……+ k)} + 1/{1 + 2 + 3 + ……+ k + (k + 1)}

= 2k/(k + 1) + 1/{1 + 2 + 3 + ……+ k + (k + 1)} [Using equation 1]

= 2k/(k + 1) + 1/{(k + 1)(k + 1 + 1)/2} [Since 1 + 2 + 3 + ……+ n = n(n + 1)/2]

= 2k/(k + 1) + 2/{(k + 1)(k + 2)}

= {2/(k + 1)}{k + 1/(k + 2)}

= {2/(k + 1)}{(k^{2} + 2k + 1)/(k + 2)}

= {2/(k + 1)}{k + 1)^{2}/(k + 2)}

= 2(k + 1)/(k + 2)

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 4:

Prove the following by using the principle of mathematical induction for all n є N:

1.2.3 + 2.3.4 + ........+ n(n + 1)(n + 2) = { n(n + 1)(n + 2)(n + 3)}/4

Answer:

Let the given statement be P(n), i.e.,

P(n): 1.2.3 + 2.3.4 + ............ + n(n + 1) (n + 2) = { n(n + 1)(n + 2)(n + 3)}/4

For n = 1, we have

P(1): 1.2.3 = 6 = {1(1 + 1)(1 + 2)(1 + 3)}/4 = (1 * 2 * 3 * 4)/6 = 6, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2.3 + 2.3.4 + ... + k(k + 1) (k + 2) = {k(k + 1)(k + 2)(k + 3)}/4 ……….1

We shall now prove that P(k + 1) is true.

Consider

1.2.3 + 2.3.4 + ... + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)

= {1.2.3 + 2.3.4 + ... + k(k + 1) (k + 2)} + (k + 1) (k + 2) (k + 3)

= {k(k + 1)(k + 2)(k + 3)}/4 + (k + 1) (k + 2) (k + 3)

= {(k + 1) (k + 2) (k + 3)}(k/4 + 1)

= {(k + 1) (k + 2) (k + 3)(k + 4)}/4

= {(k + 1) (k + 1 + 1) (k + 1 + 2)(k + 1 + 3)}/4

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 5:

Prove the following by using the principle of mathematical induction for all n є N:

1.3 + 2.3^{2} + 3.3^{3} + n.3^{n} = {(2n - 1)3^{n+1} + 3}/4

Answer:

Let the given statement be P(n), i.e.,

P(n): 1.3 + 2.3^{2} + 3.3^{3} + n3^{n} = {(2n - 1)3^{n+1} + 3}/4

For n = 1, we have

P(1): 1.3 = 3 = {(2*1 - 1)3^{1+1} + 3}/4 = (3^{2} + 3)/4 = 12/4 = 3, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.3 + 2.3^{2} + 3.3^{3} + k3^{k} = {(2k - 1)3^{k+1} + 3}/4 …………1

We shall now prove that P(k + 1) is true.

Consider

1.3 + 2.3^{2} + 3.3^{3} + ... + k.3^{k} + (k + 1).3^{k+1}

= (1.3 + 2.3^{2} + 3.3^{3} + ... + k.3^{k} ) + (k + 1).3^{k+1}

= {(2k - 1)3^{k-1} + 3}/4 + (k + 1).3^{k+1} [Using equation 1]

= {(2k - 1)3^{k-1} + 3 + 4(k + 1).3^{k+1}}/4

= [3^{k+1}{2k - 1 + 4(k + 1)} + 3]/4

= {3^{k+1}(6k + 3) + 3}/4

= {3^{k+1 } * 3(2k + 1) + 3}/4

= {3^{k+1+ 1}(2k + 1) + 3}/4

= [3^{(k+1)+ 1} {2(k + 1) – 1} + 3]/4

= [{2(k + 1) – 1}3^{(k+1)+ 1} + 3]/4

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 6:

Prove the following by using the principle of mathematical induction for all n є N:

1.2 + 2.3 + 3.4 + …………n.(n + 1) = {n(n + 1)(n + 2)}/3

Answer:

Let the given statement be P(n), i.e.,

P(n): 1.2 + 2.3 + 3.4 + …………n.(n + 1) = {n(n + 1)(n + 2)}/3

For n = 1, we have

P(1): 1.2 = 2 = {1(1 + 1)(1 + 2)}/3 = (1.2.3)/3 = 1.2 = 2, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2 + 2.3 + 3.4 + …………k.(k + 1) = {k(k + 1)(k + 2)}/3 ……………1

We shall now prove that P(k + 1) is true.

Consider

1.2 + 2.3 + 3.4 + ... + k.(k + 1) + (k + 1).(k + 2)

= [1.2 + 2.3 + 3.4 + ... + k.(k + 1)] + (k + 1).(k + 2)

= {k(k + 1)(k + 2)}/3 + (k + 1).(k + 2) [From equation 1]

= (k + 1)(k + 2)(k/3 + 1)

= {(k + 1)(k + 2)(k + 3)}/3

= {(k + 1)(k + 1 + 1)(k + 1 + 2)}/3

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 7:

Prove the following by using the principle of mathematical induction for all n є N:

1.3 + 3.5 + 5.7 + …………+ (2n - 1)(2n + 1) = n(4n^{2} + 6n - 1)/3

Answer:

Let the given statement be P(n), i.e.,

P(n): 1.3 + 3.5 + 5.7 + …………+ (2n - 1)(2n + 1) = n(4n^{2} + 6n - 1)/3

For n = 1, we have

P(1): 1.3 = 3 = 1(4.1^{2} + 6.1 - 1)/3 = (4 + 6 - 1) = 9/3 = 3, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.3 + 3.5 + 5.7 + …………+ (2k - 1)(2k + 1) = k(4k^{2} + 6k - 1)/3 ……………1

We shall now prove that P(k + 1) is true.

Consider

(1.3 + 3.5 + 5.7 + ... + (2k – 1) (2k + 1) + {2(k + 1) – 1}{2(k + 1) + 1}

= k(4k^{2} + 6k - 1)/3 + {2k + 2 – 1}{2k + 2 + 1} [From equation 1]

= k(4k^{2} + 6k - 1)/3 + (2k + 1)(2k + 3)

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

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

= (4k^{3} + 6k^{2} - k + 12k^{2} + 24k + 9)}/3

= (4k^{3} + 18k^{2} + 23k + 9)}/3

= (4k^{3} + 14k^{2} + 9k + 4k^{2} + 14k + 9)/3

= {k(4k^{2} + 14k + 9) + 1(4k^{2} + 14k + 9)}/3

= {(k + 1)(4k^{2} + 14k + 9)}/3

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

= [(k + 1){4(k^{2} + 2k + 1) + 6(k + 1) - 1}]/3

= [(k + 1){4(k + 1)^{2} + 6(k + 1) - 1}]/3

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 8:

Prove the following by using the principle of mathematical induction for all n є N:

1.2 + 2.2^{2} + 3.2^{2} + ... + n.2^{n} = (n – 1)2^{n+1} + 2

Answer:

Let the given statement be P(n), i.e.,

P(n): 1.2 + 2.2^{2} + 3.2^{2} + ... + n.2^{n} = (n – 1) 2^{n+1} + 2

For n = 1, we have

P(1): 1.2 = 2 = (1 – 1) 2^{1+1} + 2 = 0 + 2 = 2, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2 + 2.2^{2} + 3.2^{2} + ... + k.2^{k} = (k – 1) 2^{k + 1} + 2 ...........1

We shall now prove that P(k + 1) is true.

Consider

{1.2 + 2.2^{2} + 3.2^{2} + ... + k.2^{k}} + (k + 1).2^{k+1}

= (k – 1)2^{k + 1} + 2 + (k + 1).2^{k+1}

= 2^{k + 1}{(k - 1) + (k + 1)} + 2

= 2^{k + 1}.2k + 2

= k.2^{(k + 1)+1} + 2

= {(k + 1) – 1}.2^{(k + 1)+1} + 2

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 9:

Prove the following by using the principle of mathematical induction for all n є N:

1/2 + 1/4 + 1/8 + …………..+ 1/2^{n} = 1 – 1/2^{n}

Answer:

Let the given statement be P(n), i.e.,

P(n): 1/2 + 1/4 + 1/8 + …………..+ 1/2^{n} = 1 – 1/2^{n}

For n = 1, we have

P(1): 1/2 = 1 – 1/2^{1} = 1 – 1/2 = 1/2, which is true.

Let P(k) be true for some positive integer k, i.e.,

1/2 + 1/4 + 1/8 + …………..+ 1/2^{k} = 1 – 1/2^{k} …………1

We shall now prove that P(k + 1) is true.

Consider

(1/2 + 1/4 + 1/8 + …………..+ 1/2^{k}) + 1/2^{k+1}

= 1 – 1/2^{k }+ 1/2^{k+1 } [Using equation 1]

= 1 – 1/2^{k }+ 1/(2.2^{k })

= 1 – 1/2^{k }(1 – 1/2)

= 1 – 1/2^{k }(1/2)

= 1 – 1/2^{k+1}

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 10:

Prove the following by using the principle of mathematical induction for all n є N:

1/2.5 + 1/5.8 + 1/8.11 + ………….+ 1/{(3n - 1)(3n + 1)} = n/(6n + 4)

Answer:

Let the given statement be P(n), i.e.,

P(n): 1/2.5 + 1/5.8 + 1/8.11 + ………….+ 1/{(3n - 1)(3n + 1)} = n/(6n + 4)

For n = 1, we have

P(1): 1/2.5 = 1/10 = 1/(6.1 + 4) = 1/10, which is true.

Let P(k) be true for some positive integer k, i.e.,

1/2.5 + 1/5.8 + 1/8.11 + ………….+ 1/{(3k - 1)(3k + 1)} = k/(6k + 4) ……………1

We shall now prove that P(k + 1) is true.

Consider

1/2.5 + 1/5.8 + 1/8.11 + ………….+ 1/{(3k - 1)(3k + 1)} + 1/[{3(k + 1) – 1}{3(k + 1) + 2}]

= k/(6k + 4) + 1/[{3(k + 1) – 1}{3(k + 1) + 2}] [using equation 1]

= k/(6k + 4) + 1/{(3k + 3 – 1)(3k + 3 + 2)}

= k/(6k + 4) + 1/{(3k + 2)(3k + 5)}

= k/{2(3k + 2)} + 1/{(3k + 2)(3k + 5)}

= 1/(3k + 2){k/2 + 1/(3k + 5)}

= 1/(3k + 2)[{k(3k + 5) + 2}/{2(3k + 5)}]

= 1/(3k + 2)[{(3k^{2} + 5k + 2}/{6k + 10}]

= 1/(3k + 2)[{(3k + 2)(k + 1)}/{6k + 10}]

= (k + 1)/(6k + 10)

= (k + 1)/{6(k + 1) + 4}

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 11:

Prove the following by using the principle of mathematical induction for all n є N:

1/1.2.3 + 1/2.3.4 + 1/3.4.5 + …………..+ 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)}

Answer:

Let the given statement be P(n), i.e.,

P(n): 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + …………..+ 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)}

For n = 1, we have

P(1): 1/1.2.3 = {1(1 + 3)}/{4(1 + 1)(1 + 2)} = (1.4)/(4.2.3) = 1/1.2.3, which is true.

Let P(k) be true for some positive integer k, i.e.,

1/1.2.3 + 1/2.3.4 + 1/3.4.5 + …………..+ 1/{k(k + 1)(k + 2)} = {k(k + 3)}/{4(k + 1)(k + 2)}

We shall now prove that P(k + 1) is true.

Consider

1/1.2.3 + 1/2.3.4 + 1/3.4.5 + …………..+ 1/{k(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}

= {k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)} [From equation 1]

= [1/{(k + 1)(k + 2)}]{k(k + 3)/4 + 1/(k + 3)}

= [1/{(k + 1)(k + 2)}][{k(k + 3)^{2} + 4}/(k + 3)]

= [1/{(k + 1)(k + 2)}][{k(k^{2} + 6k + 9) + 4}/(k + 3)]

= [1/{(k + 1)(k + 2)}][{k^{3} + 6k^{2} + 9k + 4}/(k + 3)]

= [1/{(k + 1)(k + 2)}][{k^{3} + 2k^{2} + k + 4k^{2} + 8k + 4}/(k + 3)]

= [1/{(k + 1)(k + 2)}][{k(k^{2} + 2k + 1) + 4(k^{2} + 2k + 1)}/(k + 3)]

= [1/{(k + 1)(k + 2)}][{k(k + 1)^{2} + 4(k + 1)^{2}}/(k + 3)]

= [1/{(k + 1)(k + 2)}][{(k + 4)(k + 1)^{2}}/(k + 3)]

= {(k + 4)(k + 1)^{2}}/{4(k + 1)(k + 2)(k + 3)}

= {(k + 4)(k + 1)}/{4(k + 2)(k + 3)}

= {(k + 1)(k + 1 + 3)}/{4(k + 1 + 1)(k + 1 + 2)}

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 12:

Prove the following by using the principle of mathematical induction for all n є N:

a + ar + ar^{2} + ………..+ ar^{n-1} = a(r^{n} – 1)/(r - 1)

Answer:

Let the given statement be P(n), i.e.,

P(n): a + ar + ar^{2} + ………..+ ar^{n-1} = a(r^{n} – 1)/(r - 1)

For n = 1, we have

P(1): a = a(r^{1} – 1)/(r - 1) = a(r – 1)/(r - 1) = a, which is true.

Let P(k) be true for some positive integer k, i.e.,

a + ar + ar^{2} + ………..+ ar^{k-1} = a(r^{k} – 1)/(r - 1)

We shall now prove that P(k + 1) is true.

Consider

a + ar + ar^{2} + ………..+ ar^{k-1} + ar^{(k + 1) - 1}

= a(r^{k} – 1)/(r - 1) + ar^{k } [from equation 1]

= {a(r^{k} – 1) + ar^{k }(r - 1)} /(r - 1)

= (ar^{k} – a + ar^{k+1 }- ar^{k}) /(r - 1)

= (ar^{k+1 }- a) /(r - 1)

= a(r^{k+1 }- 1) /(r - 1)

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 13:

Prove the following by using the principle of mathematical induction for all n є N:

(1 + 3/1)(1 + 5/4)(1 + 7/9)………….. {1 + (2n + 1)/n^{2}} = (n + 1)^{2}

Answer:

Let the given statement be P(n), i.e.,

P(n): (1 + 3/1)(1 + 5/4)(1 + 7/9)………….. {1 + (2n + 1)/n^{2}} = (n + 1)^{2}

For n = 1, we have

P(1): (1 + 3/1) = 4 = (1 + 1)^{2} = 2^{2} = 4, which is true.

Let P(k) be true for some positive integer k, i.e.,

(1 + 3/1)(1 + 5/4)(1 + 7/9)………….. {1 + (2k + 1)/k^{2}} = (k + 1)^{2} …………1

We shall now prove that P(k + 1) is true.

Consider

(1 + 3/1) (1 + 5/4) (1 + 7/9)………….. {1 + (2k + 1)/k^{2}}[1 + {2(k + 1) + 1}/(k + 1)^{2}]

= (k + 1)^{2}[1 + {2(k + 1) + 1}/(k + 1)^{2}] [From equation 1]

= (k + 1)^{2}[(k + 1)^{2} + {2(k + 1) + 1}]/(k + 1)^{2}

= (k + 1)^{2} + 2(k + 1) + 1

= {(k + 1) + 1}^{2}

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 14:

Prove the following by using the principle of mathematical induction for all n є N:

(1 + 1/1)(1 + 1/2)(1 + 1/3)………….(1 + 1/n) = (n + 1)

Answer:

Let the given statement be P(n), i.e.,

P(n): (1 + 1/1)(1 + 1/2)(1 + 1/3)………….(1 + 1/n) = (n + 1)

For n = 1, we have

P(1): (1 + 1/1) = 2 = (1 + 1), which is true.

Let P(k) be true for some positive integer k, i.e.,

P(k): (1 + 1/1)(1 + 1/2)(1 + 1/3)………….(1 + 1/k) = k + 1 …………..1

We shall now prove that P(k + 1) is true.

Consider

(1 + 1/1)(1 + 1/2)(1 + 1/3)………….(1 + 1/k){1 + 1/(k + 1)}

= (k + 1) {1 + 1/(k + 1)} [From equation 1]

= (k + 1){(k + 1 + 1)/(k + 1)}

= (k + 1 + 1)

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 15:

Prove the following by using the principle of mathematical induction for all n є N:

1^{2} + 3^{2} + 5^{2} + ……………..+ (2n -1)^{2} = {n(2n - 1)(2n + 1)}/3

Answer:

Let the given statement be P(n), i.e.,

P(n): 1^{2} + 3^{2} + 5^{2} + ……………..+ (2n -1)^{2} = {n(2n - 1)(2n + 1)}/3

For n = 1, we have

P(1): 1^{2} = 1 = {1(2.1 - 1)(2.1 + 1)}/3 = (1.1.3)/3 = 1, which is true.

Let P(k) be true for some positive integer k, i.e.,

P(k): 1^{2} + 3^{2} + 5^{2} + ……………..+ (2k -1)^{2} = {k(2k - 1)(2k + 1)}/3 ………….1

We shall now prove that P(k + 1) is true.

Consider

1^{2} + 3^{2} + 5^{2} + ……………..+ (2k -1)^{2} + {2(k + 1) - 1}^{2}

= {k(2k - 1)(2k + 1)}/3 + {2(k + 1) - 1}^{2 } [From equation 1]

= [k(2k - 1)(2k + 1) + 3{2(k + 1) - 1}^{2}]/3

= [k(2k - 1)(2k + 1) + 3{2k + 2 - 1}^{2}]/3

= [k(2k - 1)(2k + 1) + 3(2k + 1)^{2}]/3

= (2k + 1)[k(2k - 1) + 3(2k + 1)]/3

= (2k + 1)[2k^{2} - k + 6k + 3]/3

= (2k + 1)[2k^{2} + 5k + 3]/3

= (2k + 1)[2k^{2} + 2k + 3k + 3]/3

= (2k + 1)[2k(k + 1) + 3(k + 1)]/3

= {(2k + 1)(k + 1)(2k + 3)}/3

= [(k + 1){ (2(k + 1) - 1}(2(k + 1) + 1}]/3

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 16:

Prove the following by using the principle of mathematical induction for all n є N:

1/1.4 + 1/4.7 + 1/7.10 + ………….+ 1/{(3n - 2)(3n + 1)} = n/(3n + 1)

Answer:

Let the given statement be P(n), i.e.,

P(n): 1/1.4 + 1/4.7 + 1/7.10 + ………….+ 1/{(3n - 2)(3n + 1)} = n/(3n + 1)

For n = 1, we have

P(1) = 1/1.4 = 1/(3.1 + 1) = 1/4 = 1/1.4, which is true.

Let P(k) be true for some positive integer k, i.e.,

P(k): 1/1.4 + 1/4.7 + 1/7.10 + ………….+ 1/{(3k - 2)(3k + 1)} = k/(3k + 1) ………..1

We shall now prove that P(k + 1) is true.

Consider

1/1.4 + 1/4.7 + 1/7.10 + ………….+ 1/{(3k - 2)(3k + 1)} + 1/[{3(k + 1) – 2}{3(k + 1) + 1)}]

= k/(3k + 1) + 1/[{3(k + 1) – 2}{3(k + 1) + 1)}] [From equation 1]

= k/(3k + 1) + 1/{(3k + 1)(3k + 4)}

= {1/(3k + 1)}{k + 1/(3k + 4)}

= {1/(3k + 1)}[{k(3k + 4) + 1}/(3k + 4)]

= {1/(3k + 1)}{3k^{2} + 4k + 1}/(3k + 4)

= {1/(3k + 1)}{3k^{2} + 3k + k + 1}/(3k + 4)

= {1/(3k + 1)}{(3k + 1)(k + 1)}/(3k + 4)

= (k + 1)/(3k + 4)

= (k + 1)/{(3(k + 1) + 1}

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 17:

Prove the following by using the principle of mathematical induction for all n є N:

1/3.5 + 1/5.7 + 1/7.9 + ………….+ 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3)}

Answer:

Let the given statement be P(n), i.e.,

P(n): 1/3.5 + 1/5.7 + 1/7.9 + ………….+ 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3)}

For n = 1, we have

P(1): 1/3.5 = 1/{3(2.1 + 3)} = 1/3.5, which is true.

Let P(k) be true for some positive integer k, i.e.,

P(k): 1/3.5 + 1/5.7 + 1/7.9 + ………….+ 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} …………1

We shall now prove that P(k + 1) is true.

Consider

1/3.5 + 1/5.7 + 1/7.9 + ………….+ 1/{(2k + 1)(2k + 3)} + 1/[{2(k + 1) + 1}{(2(k + 1) + 3}]

= k/{3(2k + 3)} + 1/[{2(k + 1) + 1}{(2(k + 1) + 3}] [From equation 1]

= k/{3(2k + 3)} + 1/{(2k + 3)(2k + 5)}

= {1/(2k + 3)}{k/3 + 1/(2k + 5)}

= {1/(2k + 3)}[{k(2k + 5) + 3}/{3(2k + 5)}]

= {1/(2k + 3)}{2k^{2} + 5k + 3}/{3(2k + 5)}

= {1/(2k + 3)}{2k^{2} + 2k + 3k + 3}/{3(2k + 5)}

= {1/(2k + 3)}{2k(k + 1) + 3(k + 1)}/ {3(2k + 5)}

= {1/(2k + 3)}{(2k + 3)(k + 1)}/ {3(2k + 5)}

= (k + 1)}/{3(2k + 5)}

= (k + 1)}/[3{2(k + 1) + 3)}]

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 18:

Prove the following by using the principle of mathematical induction for all n є N:

1 + 2 + 3 + …………+ n < (2n + 1)^{2}/8

Answer:

Let the given statement be P(n), i.e.,

P(n): 1 + 2 + 3 + …………+ n < (2n + 1)^{2}/8

It can be noted that P(n) is true for n = 1 since

1 < (2.1 + 1)^{2}/8 = 9/8

Let P(k) be true for some positive integer k, i.e.,

P(k): 1 + 2 + 3 + …………+ k < (2k + 1)^{2}/8 ……….1

We shall now prove that P(k + 1) is true.

Consider

1 + 2 + 3 + …………+ k + (k + 1) < (2k + 1)^{2}/8 + (k + 1) [from equation 1]

1 + 2 + 3 + …………+ k + (k + 1) < {(2k + 1)^{2} + 8(k + 1)}/8

1 + 2 + 3 + …………+ k + (k + 1) < {4k^{2} + 4k + 1 + 8k + 8}/8

1 + 2 + 3 + …………+ k + (k + 1) < {4k^{2} + 12k + 9}/8

1 + 2 + 3 + …………+ k + (k + 1) < (2k + 3)^{2}/8

1 + 2 + 3 + …………+ k + (k + 1) < {2(k + 1) + 1}^{2}/8

Thus, 1 + 2 + 3 + …………+ k + (k + 1) < (2k + 1)^{2}/8 + (k + 1)

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 19:

Prove the following by using the principle of mathematical induction for all n ∈ N:

n(n + 1)(n + 5) is a multiple of 3.

Answer:

Let the given statement be P(n), i.e.,

P(n): n (n + 1) (n + 5), which is a multiple of 3.

It can be noted that P(n) is true for n = 1

Since 1(1 + 1)(1 + 5) = 1 * 2 * 6 = 12, which is a multiple of 3.

Let P(k) be true for some positive integer k, i.e.,

k(k + 1)(k + 5) is a multiple of 3.

So, k(k + 1)(k + 5) = 3m, where m ∈ N ...........1

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

(k + 1){(k + 1) + 1}{(k + 1) + 5}

= (k + 1)(k + 2){(k + 5) + 1}

= (k + 1)(k + 2)(k + 5) + (k + 1)(k + 2)

= k(k + 1)(k + 5) + 2(k + 1)(k + 5) + (k + 1)(k + 2)

= 3m + (k + 1){2(k + 5) + (k + 2)}

= 3m + (k + 1){2k + 10 + k + 2}

= 3m + (k + 1)(3k + 12)

= 3m + 3(k + 1)(k + 4)

= 3{m + (k + 1)(k + 4)}

= 3 * q, where q = m + (k + 1)(k + 4) is some natural number.

Therefore, (k + 1){(k + 1) + 1}{(k + 1) + 5} is a multiple of 3

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 20:

Prove the following by using the principle of mathematical induction for all n ∈ N:

10^{2n – 1} + 1 is divisible by 11.

Answer:

Let the given statement be P(n), i.e.,

P(n): 10^{2n – 1} + 1 is divisible by 11.

It can be observed that P(n) is true for n = 1

Since P(1) = 10^{2.1 – 1} + 1 = 11, which is divisible by 11.

Let P(k) be true for some positive integer k,

i.e., 10^{2k – 1} + 1 is divisible by 11.

So, 10^{2k – 1} + 1 = 11m, where m ∈ N ..............1

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

10^{2(k + 1) - 1} + 1

= 10^{2k + 2 - 1} + 1

= 10^{2 }(10^{2k + 2 - 1} + 1 - 1) + 1

= 10^{2 }(10^{2k + 2 - 1} + 1) - 10^{2} + 1

= 10^{2 }. 11m – 100 + 1

= 100 * 11m – 100 + 1

= 100 * 11m – 99

= 11(100m – 9)

= 11r, where r = (100m - 9) is some natural number.

Therefore, 10^{2(k + 1) - 1} + 1is divisible by 11.

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 21:

Prove the following by using the principle of mathematical induction for all n ∈ N:

x^{2n} – y^{2n} is divisible by x + y.

Answer:

Let the given statement be P(n), i.e.,

P(n): x^{2n} – y^{2n} is divisible by x + y

It can be observed that P(n) is true for n = 1.

This is so because x^{2 * 1} – y^{2 * 1} = x^{2} – y^{2} = (x + y)(x – y) is divisible by (x + y).

Let P(k) be true for some positive integer k, i.e.,

x^{2k} – y^{2k} is divisible by x + y.

Let x^{2k} – y^{2k} = m(x + y), where m ∈ N ..........1

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

x^{2(k + 1)} – y^{2(k + 1)}

= x^{2k} * x^{2} – y^{2k} * y^{2}

= x^{2k} * x^{2} – y^{2k} * x^{2} + y^{2k} * x^{2} - y^{2k} * y^{2}

= x^{2}(x^{2k} – y^{2k} + y^{2k}) - y^{2k} * y^{2}

= x^{2}{m(x + y) + y^{2k}} - y^{2k} * y^{2 } [From equation 1]

= m(x + y)x^{2} + y^{2k} * x^{2} - y^{2k} * y^{2}

= m(x + y)x^{2} + y^{2k}(x^{2} - y^{2})

= m(x + y)x^{2} + y^{2k}(x - y)(x + y)

= (x + y){mx^{2} + y^{2k}(x - y)}, which is a factor of (x + y).

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 22:

Prove the following by using the principle of mathematical induction for all n ∈ N:

3^{2n + 2} – 8n – 9 is divisible by 8.

Answer:

Let the given statement be P(n), i.e.,

P(n): 3^{2n + 2} – 8n – 9 is divisible by 8.

It can be observed that P(n) is true for n = 1

Since 3^{2 * 1} + 2 – 8 * 1 – 9 = 64, which is divisible by 8.

Let P(k) be true for some positive integer k, i.e.,

3^{2k + 2} – 8k – 9 is divisible by 8.

3^{2k + 2} – 8k – 9 = 8m; where m ∈ N ..............1

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

3^{2(k + 1) + 2} – 8(k + 1) – 9

= 3^{2k + 2 + 2} – 8k - 8 – 9

= 3^{2k + 2 }* 3^{2} – 8k – 9

= 3^{2}(3^{2k + 2} – 8k – 9 + 8k + 9) – 8k - 17

= 3^{2}(3^{2k + 2} – 8k – 9) + 3^{2}(8k + 9) – 8k - 17

= 9 * 8m + 9(8k + 9) – 8k - 17

= 9 * 8m + 72k + 81 – 8k - 17

= 9 * 8m + 64k + 64

= 8(9m + 8k + 8)

= 8r, where r = (9m + 8k + 8) is a natural number.

Therefore, 3^{2(k + 1) + 2} – 8(k + 1) – 9 is divisible by 8.

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 23:

Prove the following by using the principle of mathematical induction for all n ∈ N:

41^{n} – 14^{n }is a multiple of 27.

Answer:

Let the given statement be P(n), i.e.,

P(n): 41^{n} – 14^{n} is a multiple of 27.

It can be observed that P(n) is true for n = 1

Since 41^{1} – 14^{1} = 27, which is a multiple of 27.

Let P(k) be true for some positive integer k, i.e.,

41^{k} – 14^{k} is a multiple of 27

So, 41^{k} – 14^{k} = 27m, where m ∈ N .............1

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

41^{k+1} – 14^{k+1}

= 41^{k} * 41 – 14^{k }* 14

= 41(41^{k} – 14^{k }+ 14^{k}) - 14^{k }+ 14

= 41(41^{k} – 14^{k }) + 41 * 14^{k} - 14^{k }+ 14

= 41 * 27m + 14^{k}(41 – 14)

= 41 * 27m + 27 * 14^{k}

= 27(41m + 14^{k})

= 27 * r, where r = (41m + 14^{k}) is a natural number.

Therefore, 41^{k+1} – 14^{k+1} is a multiple of 27

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

Question 24:

Prove the following by using the principle of mathematical induction for all n є N:

(2n +7) < (n + 3)^{2}

Answer:

Let the given statement be P(n), i.e.,

P(n): (2n +7) < (n + 3)^{2}

It can be observed that P(n) is true for n = 1

Since 2.1 + 7 = 9 < (1 + 3)^{2} = 16, which is true.

Let P(k) be true for some positive integer k, i.e.,

(2k + 7) < (k + 3)^{2} ............1

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider 2(k + 1) + 7 = (2k + 7) + 2

So, 2(k + 1) + 7 = (2k + 7) + 2 < (k + 3)^{2} + 2 [From equation 1]

= 2(k + 1) + 7 < k^{2} + 6k + 9 + 2

= 2(k + 1) + 7 < k^{2} + 6k + 11

Now, k^{2} + 6k + 11 < k^{2} + 8k + 16

So, 2(k + 1) + 7 < (k + 4)^{2}

= 2(k + 1) + 7 < {(k + 1) + 3}^{2}

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., N.

.