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 + 32 + 33 + …… + 3n-1 = (3n – 1)/2

Answer 1:

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

P(n): 1 + 3 + 32 + 33 + …… + 3n-1 = (3n – 1)/2

For n = 1, we have

P(1):= (31 – 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 + 32 + 33 + …… + 3k-1 = (3k – 1)/2…………..1

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

Consider

1 + 3 + 32 + ... + 3k–1 + 3(k+1) – 1 = (1 + 3 + 32 +... + 3k–1) + 3k

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

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

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

                                                    = {3 * 3k – 1}/2

                                                    = (3k+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:

13 + 23 + 33 + ………..+ n3 = {n(n + 1)/2}2

Answer:

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

P(n): 13 + 23 + 33 + ………..+ n3 = {n(n + 1)/2}2

For n = 1, we have

P(1): 13 =1= {1(1 + 1)/2}2 = {(1 * 2)/2}2 = 12 = 1, which is true.

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

13 + 23 + 33 + ………..+ k3 = {k(k + 1)/2}2   ……………..1

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

Consider

13 + 23 + 33 + ... + k3 + (k + 1)3

= (13 + 23 + 33 + .... + k3) + (k + 1)3

= {k(k + 1)/2}2 + (k + 1)3 

= k2(k + 1)2/4 + (k + 1)3

= {k2(k + 1)2 + 4(k + 1)3 }/4

= (k + 1)2{k2 + 4(k + 1)}/4

= (k + 1)2{k2 + 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)}{(k2 + 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 + 2.32 + 3.33 + n.3n = {(2n - 1)3n+1 + 3}/4

Answer:

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

P(n): 1.3 + 2.32 + 3.33 + n3n = {(2n - 1)3n+1 + 3}/4

For n = 1, we have

P(1): 1.3 = 3 = {(2*1 - 1)31+1 + 3}/4 = (32 + 3)/4 = 12/4 = 3, which is true.

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

1.3 + 2.32 + 3.33 + k3k = {(2k - 1)3k+1 + 3}/4  …………1

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

Consider

    1.3 + 2.32 + 3.33 + ... + k.3k + (k + 1).3k+1

= (1.3 + 2.32 + 3.33 + ... + k.3k ) + (k + 1).3k+1

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

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

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

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

= {3k+1  * 3(2k + 1) + 3}/4

= {3k+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(4n2 + 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(4n2 + 6n - 1)/3

For n = 1, we have

P(1): 1.3 = 3 = 1(4.12 + 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(4k2 + 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(4k2 + 6k - 1)/3 + {2k + 2 – 1}{2k + 2 + 1}            [From equation 1]

= k(4k2 + 6k - 1)/3 + (2k + 1)(2k + 3)

= k(4k2 + 6k - 1)/3 + (4k2 + 8k + 3)

= {k(4k2 + 6k - 1) + 3(4k2 + 8k + 3)}/3

= (4k3 + 6k2 - k + 12k2 + 24k + 9)}/3

= (4k3 + 18k2 + 23k + 9)}/3

= (4k3 + 14k2 + 9k + 4k2 + 14k + 9)/3

= {k(4k2 + 14k + 9) + 1(4k2 + 14k + 9)}/3

= {(k + 1)(4k2 + 14k + 9)}/3

= {(k + 1)(4k2 + 8k + 4 + 6k + 6 - 1)}/3

= [(k + 1){4(k2 + 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 + 2.22 + 3.22 + ... + n.2n = (n – 1)2n+1 + 2

Answer:

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

P(n): 1.2 + 2.22 + 3.22 + ... + n.2n = (n – 1) 2n+1 + 2

For n = 1, we have

P(1): 1.2 = 2 = (1 – 1) 21+1 + 2 = 0 + 2 = 2, which is true.

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

1.2 + 2.22 + 3.22 + ... + k.2k = (k – 1) 2k + 1 + 2    ...........1

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

Consider

    {1.2 + 2.22 + 3.22 + ... + k.2k} + (k + 1).2k+1

= (k – 1)2k + 1 + 2 + (k + 1).2k+1

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

= 2k + 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 + …………..+ 1/2n = 1 – 1/2n

Answer:

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

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

For n = 1, we have

P(1): 1/2 = 1 – 1/21 = 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 + …………..+ 1/2k = 1 – 1/2k   …………1

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

Consider

    (1/2 + 1/4 + 1/8 + …………..+ 1/2k) + 1/2k+1

= 1 – 1/2k + 1/2k+1                [Using equation 1]

= 1 – 1/2k + 1/(2.2k )

= 1 – 1/2k (1 – 1/2)

= 1 – 1/2k (1/2)

= 1 – 1/2k+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)[{(3k2 + 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(k2 + 6k + 9) + 4}/(k + 3)]

= [1/{(k + 1)(k + 2)}][{k3 + 6k2 + 9k + 4}/(k + 3)]

= [1/{(k + 1)(k + 2)}][{k3 + 2k2 + k + 4k2 + 8k + 4}/(k + 3)]

= [1/{(k + 1)(k + 2)}][{k(k2 + 2k + 1) + 4(k2 + 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 + ar2 + ………..+ arn-1 = a(rn – 1)/(r - 1)

Answer:

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

P(n): a + ar + ar2 + ………..+ arn-1 = a(rn – 1)/(r - 1)

For n = 1, we have

P(1): a = a(r1 – 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 + ar2 + ………..+ ark-1 = a(rk – 1)/(r - 1)

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

Consider

    a + ar + ar2 + ………..+ ark-1 + ar(k + 1) - 1

= a(rk – 1)/(r - 1) + ark               [from equation 1]

= {a(rk – 1) + ark (r - 1)} /(r - 1)

= (ark – a + ark+1 - ark) /(r - 1)

= (ark+1 - a) /(r - 1)

= a(rk+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)/n2} = (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)/n2} = (n + 1)2

For n = 1, we have

P(1): (1 + 3/1) = 4 = (1 + 1)2 = 22 = 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)/k2} = (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)/k2}[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:

12 + 32 + 52 + ……………..+ (2n -1)2 = {n(2n - 1)(2n + 1)}/3

Answer:

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

P(n): 12 + 32 + 52 + ……………..+ (2n -1)2 = {n(2n - 1)(2n + 1)}/3

For n = 1, we have

P(1): 12 = 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): 12 + 32 + 52 + ……………..+ (2k -1)2 = {k(2k - 1)(2k + 1)}/3   ………….1

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

Consider

    12 + 32 + 52 + ……………..+ (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)[2k2 - k + 6k + 3]/3

= (2k + 1)[2k2 + 5k + 3]/3

= (2k + 1)[2k2 + 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)}{3k2 + 4k + 1}/(3k + 4)

= {1/(3k + 1)}{3k2 + 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)}{2k2 + 5k + 3}/{3(2k + 5)}

= {1/(2k + 3)}{2k2 + 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) < {4k2 + 4k + 1 + 8k + 8}/8

1 + 2 + 3 + …………+ k + (k + 1) < {4k2 + 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:

102n – 1 + 1 is divisible by 11.

Answer:

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

P(n): 102n – 1 + 1 is divisible by 11.

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

Since P(1) = 102.1 – 1 + 1 = 11, which is divisible by 11.

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

i.e., 102k – 1 + 1 is divisible by 11.

So, 102k – 1 + 1 = 11m, where m ∈ N ..............1

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

Consider

    102(k + 1) - 1 + 1

= 102k + 2 - 1 + 1

= 102 (102k + 2 - 1 + 1 - 1) + 1

= 102 (102k + 2 - 1 + 1) - 102 + 1

= 102 . 11m – 100 + 1

= 100 * 11m – 100 + 1

= 100 * 11m – 99

= 11(100m – 9)

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

Therefore, 102(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:

x2n – y2n is divisible by x + y.

Answer:

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

P(n): x2n – y2n is divisible by x + y

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

This is so because x2 * 1 – y2 * 1 = x2 – y2 = (x + y)(x – y) is divisible by (x + y).

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

x2k – y2k is divisible by x + y.

Let x2k – y2k = m(x + y), where m ∈ N   ..........1

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

Consider

    x2(k + 1) – y2(k + 1)

= x2k * x2 – y2k * y2

= x2k * x2 – y2k * x2 + y2k * x2 - y2k * y2

= x2(x2k – y2k + y2k) - y2k * y2

= x2{m(x + y) + y2k} - y2k * y2              [From equation 1]

= m(x + y)x2 + y2k * x2 - y2k * y2

= m(x + y)x2 + y2k(x2 - y2)

= m(x + y)x2 + y2k(x - y)(x + y)

= (x + y){mx2 + y2k(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:

32n + 2 – 8n – 9 is divisible by 8.

Answer:

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

P(n): 32n + 2 – 8n – 9 is divisible by 8.

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

Since 32 * 1 + 2 – 8 * 1 – 9 = 64, which is divisible by 8.

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

32k + 2 – 8k – 9 is divisible by 8.

32k + 2 – 8k – 9 = 8m; where m ∈ N   ..............1

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

Consider

   32(k + 1) + 2 – 8(k + 1) – 9

= 32k + 2 + 2 – 8k - 8 – 9

= 32k + 2 * 32 – 8k – 9

= 32(32k + 2 – 8k – 9 + 8k + 9) – 8k - 17

= 32(32k + 2 – 8k – 9) + 32(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, 32(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:

41n – 14n is a multiple of 27.

Answer:

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

P(n): 41n – 14n is a multiple of 27.

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

Since 411 – 141 = 27, which is a multiple of 27.

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

41k – 14k is a multiple of 27

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

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

Consider

    41k+1 – 14k+1

= 41k * 41 – 14k * 14

= 41(41k – 14+ 14k) - 14+ 14

= 41(41k – 14k ) + 41 * 14k - 14+ 14

= 41 * 27m + 14k(41 – 14)

= 41 * 27m + 27 * 14k

= 27(41m + 14k)

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

Therefore, 41k+1 – 14k+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 < k2 + 6k + 9 + 2

= 2(k + 1) + 7 < k2 + 6k + 11

Now, k2 + 6k + 11 < k2 + 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.

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