Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Mathematical Induction Solutions Exercise 2(a) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1A Mathematical Induction Solutions Exercise 2(a)

Using mathematical induction, prove each of the following statements for all n ∈ N.

Question 1.

1^{2} + 2^{2} + 3^{2} + …… + n^{2} = \(\frac{n(n+1)(2 n+1)}{6}\)

Solution:

Let p(n) be the given statement:

1^{2} + 2^{2} + 3^{2} + ….. + n^{2} = \(\frac{n(n+1)(2 n+1)}{6}\)

Since 1^{2} = \(\frac{(1)(1+1)(2 \times 1+1)}{6}\)

⇒ 1 = 1 the formula is true for n = 1

i.e., p(1) is true.

Assume the statement p(n) is true for n = k

i.e., 1^{2} + 2^{2} + 3^{2} + …… + 1k^{2} = \(\frac{k(k+1)(2 k+1)}{6}\)

We show that the formula is true for n = k + 1

i.e., We show that p(k + 1) = \(\frac{(k+1)(k+2)(2 k+3)}{6}\)

(Where p(k) = 1^{2} + 2^{2} + 3^{2} + … + k^{2})

We observe that

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

Since p(k) = \(\frac{k(k+1)(2 k+1)}{6}\)

We have p(k + 1) = p(k) + (k + 1)^{2}

∴ The formula holds for n = k + 1

∴ By the principle of mathematical induction, p(n) is true for all n ∈ N

i.e., the formula 1^{2} + 2^{2} + 3^{2} + ……. + n^{2} = \(\frac{n(n+1)(2 n+1)}{6}\) for all n ∈ N

Question 2.

2.3 + 3.4 + 4.5 + …… up to n terms = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)

Solution:

The nth term in the given series is (n + 1) (n + 2)

Let p(n) be the statement:

2.3 + 3.4 + 4.5 + …… + (n + 1) (n + 2) = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)

and let S(n) be the sum on the left-hand side.

Since S(1) = 2.3 = \(\frac{(1)(1+6+11)}{3}\) = 6

∴ The statement is true for n = 1

Assume that the statement p(n) is true for n = k

i.e., S(k) = 2.3 + 3.4 + …… + (k + 1) (k + 2) = \(\frac{k\left(k^{2}+6 k+11\right)}{3}\)

We show that the statement is true for n = k + 1

i.e., We show that S(k + 1) = \((k+1)\left[\frac{(k+1)^{2}+6(k+1)+11}{3}\right]\)

We observe that

S(k + 1) = 2.3 + 3.4 + 4.5 + + (k + 1) (k + 2) + (k + 2) (k + 3)

= S(k) + (k + 2) (k + 3)

∴ The statement holds for n = k + 1

∴ By the principle of mathematical induction,

p(n) is true for all n ∈ N

i.e., 2.3 + 3.4 + 4.5 + ……. + (n + 1) (n + 2) = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)

Question 3.

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)

Solution:

Let p(n) be the statement:

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)

and let S(n) be the sum on the L.H.S.

Since S(1) = \(\frac{1}{1.3}=\frac{1}{1(2+1)}=\frac{1}{1.3}\)

∴ P(1) is true.

Assume that the statement p(n) is true for n = k

i.e., S(k) = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 k-1)(2 k+1)}\) = \(\frac{k}{2 k+1}\)

We show that the statement p(n) is true for n = k + 1

i.e., we show that s(k + 1) = \(\frac{k+1}{2(k+1)+1}\)

We observe that

∴ The statement holds for n = k + 1

∴ By the principle of mathematical induction,

p(n) is true for all n ∈ N

i.e., \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)

Question 4.

4^{3} + 8^{3} + 12^{3} + …… up to n terms = 16n^{2}(n + 1)^{2}.

Solution:

4, 8, 12,….. are in A.P., whose nth term is (4n)

Let p(n) be the statement:

4^{3} + 8^{3} + 12^{3} + ………. + (4n)^{3} = 16n^{2}(n + 1)^{2}

and S(n) be the sum on the L.H.S.

S(1) = 4^{3} = 16(1^{2}) (1 + 1)^{2} = 16(4) = 64 = 4^{3}

∴ p(1) is true

Assume that the statement p(n) is true for n = k

i.e., S(k) = 4^{3} + 8^{3} + (12)^{3} + …… + (4k)^{3} = 16k^{2}(k + 1)^{2}

We show that the statement is true for n = k + 1

i.e., We show that S(k + 1) = 16(k + 1)^{2} (k + 2)^{2}

We observe that

S(k + 1) = 4^{3} + 8^{3} + 12^{3} + …… + (4k)^{3} + [4(k + 1)]^{3}

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

= 16k^{2} (k + 1)^{2} + 4^{3} (k + 1)^{3}

= 16(k + 1)^{2} [k^{2} + 4(k + 1)]

= 16(k + 1)^{2} [k^{2} + 4k + 4]

= 16(k + 1)^{2} (k + 2)^{2}

= 16(k + 1)^{2} \((\overline{k+1}+1)^{2}\)

∴ The formula holds for n = k + 1

∴ By the principle of mathematical induction,

p(n) is true for all n ∈ N

(i.e.,) 4^{3} + 8^{3} + 12^{3} + …… + (4n)^{3} = 16n^{2}(n + 1)^{2}

Question 5.

a + (a + d) + (a + 2d) + ……. up to n terms = \(\frac{n}{2}\) [2a + (n – 1)d]

Solution:

Let p(n) be the statement:

a + (a + d) + (a + 2d) + …… + [a + (n – 1)d] = \(\frac{n}{2}\) [2a + (n – 1)d]

and let the sum on the L.H.S. is denoted by S(n)

Since S(1) = a = \(\frac{1}{2}\) [2a + (1 – 1)d] = a

∴ p(1) is true.

Assume that the statement is true for n = k

(i.e.,) S(k) = a + (a + d) + (a + 2d) + ……. + [a + (k – 1)d] = \(\frac{k}{2}\) [2a + (k – 1 )d]

We show that the statement is true for n = k + 1

(i.e.,) we show that S(k + 1) = \(\left(\frac{k+1}{2}\right)[2 a+k d]\)

We observe that

S(k + 1) = a + (a + d) + (a + 2d) + …… + [a + (k – 1)d] + (a + kd)

= S(k) + (a + kd)

= \(\frac{k}{2}\) [2a + (k – 1)d] + (a + kd)

= \(\frac{k[2 a+(k-1) d]+2(a+k d)}{2}\)

= \(\frac{1}{2}\) [2ak + k(k – 1)d + 2a + 2kd]

= \(\frac{1}{2}\) [2a(k + 1) + k(k – 1 + 2)d]

= \(\frac{1}{2}\) (k + 1)(2a + kd)

∴ The statement holds for n = k + 1

∴ By the principle of mathematical inductions,

p(n) is true for all n ∈ N

(i.e.,) a + (a + d) + (a + 2d) + …… + [a + (n – 1)d] = \(\frac{n}{2}\) [2a + (n – 1)d]

Question 6.

a + ar + ar^{2} + ……… up to n terms = \(\frac{a\left(r^{n}-1\right)}{r-1}\); r ≠ 1

Solution:

Let p(n) be the statement:

a + ar + a.r^{2} + …… + a. r^{n-1} = \(\frac{a\left(r^{n}-1\right)}{r-1}\), r ≠ 1

and let S(n) be the sum on the L.H.S

Since S(1) = a = \(\frac{a\left(r^{1}-1\right)}{r-1}\) = a

∴ p(1) is true

Assume that the statement is true for n = k

(i.e) S(k) = a + ar + ar^{2} + ……… + a . r^{k-1} = \(\frac{a\left(r^{k}-1\right)}{r-1}\)

We show that the statement is true for n = k + 1

(i.e) S(k + 1) = \(\frac{a\left(r^{k+1}-1\right)}{r-1}\)

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

= S(k) + a . r^{k}

∴ The statement holds for n = k + 1

∴ By the principle of mathematical induction,

p(n) is true for all n ∈ N

(i.e) a + ar + ar^{2} + ……. + a.r^{n-1} = \(\frac{a\left(r^{n}-1\right)}{r-1}\), r ≠ 1

Question 7.

2 + 7 + 12 + ……. + (5n – 3) = \(\frac{n(5 n-1)}{2}\)

Solution:

Let p(n) be the statement:

2 + 7 + 12 + ……. + (5n – 3) = \(\frac{n(5 n-1)}{2}\)

and let S(n) be the sum on the L.H.S

Since S(1) = 2 = \(\frac{1(5 \times 1-1)}{2}=\frac{4}{2}\) = 2

∴ p(1) is true

Assume that the statement is true for n = k

(i.e) S(k) = 2 + 7 + 12 + …….. + (5k – 3) = \(\frac{k(5 k-1)}{2}\)

We have to show that S(k + 1) = \(\frac{(k+1)(5 k+4)}{2}\)

We observe that S(k + 1) = 2 + 7 + 12 + ……. + (5k – 3) + (5k + 2)

= S(k) + (5k + 2)

= \(\frac{k(5 k-1)}{2}\) + (5k + 2)

= \(\frac{5 k^{2}-k+2(5 k+2)}{2}\)

= \(\frac{1}{2}\) [5k^{2} + 9k + 4]

= \(\frac{1}{2}\) (k + 1) (5k + 4)

= \(\frac{1}{2}\) (k + 1) [5(k + 1) – 1]

∴ p(k + 1) is true

∴ By the principle of mathematical induction,

p(n) is true for all n ∈ N.

(i.e.,) 2 + 7 + 12 + …… + (5n – 3) = \(\frac{n(5 n-1)}{2}\)

Question 8.

\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\frac{2 n+1}{n^{2}}\right)\) = (n + 1)^{2}

Solution:

Let p(n) be the statement:

\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\frac{2 n+1}{n^{2}}\right)\) = (n + 1)^{2}

and let S(n) be the product on the LHS

since S(1) = 1 + 3 = 4 = (1 + 1)^{2} = 4

∴ P(a) 4 time for n = 1

Assume that p(n) is true for n = k

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

= k^{2} + 2k + 1 + 2k + 3

= k^{2} + 4k + 4

= (k + 2)^{2}

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

∴ P(n) is true for n = k + 1

By the principle of Mathematical Induction,

p(n) is true & n ∈ N

Question 9.

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

Solution:

Let p (n) be the statement

When n = 1, 9 < 16

∴ p(n) is true for n = 1

Assume p (n) is true for n = k

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

We show that p(n) is true for n = k + 1

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

< (k + 3)^{2} + 2

< k^{2} + 6k + 9 + 2 + 2k + 5 – 2k – 5

< (k + 4)^{2} – (2k + 5)

< (k + 4)^{2}

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

∴ p(n) is true for n = k + 1

By the principle of Mathematical Induction

p(n) is true ∀ n ∈ N

Question 10.

1^{2} + 2^{2} + …… + n^{2} > \(\frac{n^{3}}{3}\)

Solution:

Let P(n) by the statement

when n = 1, 1 > \(\frac{1}{3}\)

∴ p(n) is true for n = 1

Assume p (n) is true for n = k

1^{2} + 2^{2} + …… + k^{2} > \(\frac{k^{3}}{3}\)

We show that p(n) is true for n = k + 1

∴ p(n) is true for n = k + 1

By the principle of Mathematical Induction,

p(n) is true ∀ n ∈ N

Question 11.

4^{n} – 3n – 1 is divisible by 9.

Solution:

Let p(n) be the statement:

4^{n} – 3n – 1 is divisible by 9

Since 4^{1} – 3(1) – 1 = 0 is divisible by 9.

The statement is true for n = 1

Assume that p(n) is true for n = k

(i.e) 4^{k} – 3k – 1 is divisible by 9

Then 4^{k} – 3k – 1 = 9t, for some t ∈ N ……..(1)

Show that the statement p(n) is true for n = k + 1

(i.e.,) we show that S(k + 1) = 4^{k+1} – 3(k+1) – 1 is divisible by 9

From (1), we have

4^{k} = 9t + 3k + 1

∴ S(k + 1) = 4 . 4^{k} – 3(k + 1) – 1

= 4(9t + 3k + 1) – 3k – 3 – 1

= 4(9t) + 9k

= 9[4t + k]

Hence s(k + 1) is divisible by 9

Since 4t + k is an integer

∴ 4^{k+1} – 3(k+1) – 1 is divisible by 9

∴ The statement is true for n = k + 1

∴ By the principle of mathematical induction,

p(n) is true for all n ∈ k

(i.e.,) 4^{n} – 3n – 1 is divisible by 9

Question 12.

3 . 5^{2n+1} + 2^{3n+1} is divisible by 17.

Solution:

Let p(n) be the statement:

3. 5^{2n+1} + 2^{3n+1} is divisible by 17

Since 3 . 5^{2(1)+1} + 2^{3(1)+1}

= 3 . 5^{3} + 2^{4}

= 3(125) + 16

= 375 + 16

= 391

= 17(23) is divisible by 17

∴ The statement is true for n = 1

Assume that the statement is true for n = k

(i.e) 3 . 5^{2k+1} + 2^{3k+1} is divisible by 17

Then 3 . 5^{2k+1} + 2^{3k+1} = 17t, for some t ∈ N ……..(1)

Show that the statement p(n) is true for n = k + 1

(i.e.,) We have to show that

\(\text { 3. } 5^{2(k+1)+1}+2^{3(k+1)+1}\) is divisible by 17

From (1) we have

Here 25t + 2^{3k+1} is an integer

∴ \(\text { 3. } 5^{2(k+1)+1}+2^{3(k+1)+1}\) is divisible by 17

∴ The statement is true for n = k + 1

∴ By the principle of mathematical induction,

p(n) is true for all n ∈ N

(i.e.,) 3 . 5^{2n+1} + 2^{3n+1} is divisible by 17.

Question 13.

1.2.3 + 2.3.4 + 3.4.5 + ……. upto n terms = \(\frac{n(n+1)(n+2)(n+3)}{4}\)

Solution:

The nth term of the given series is (n) (n + 1) (n + 2)

Let p(n) be the statement:

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

and S(n) be the sum on the L.H.S.

∵ S(1) = 1.2.3 = \(\frac{(1)(1+1)(1+2)(1+3)}{4}\) = 1.2.3

∴ p(1) is true

Assume that the statement p(n) is true for n = k

(i.e) S(k) = 1.2.3 + 2.3.4 + 3.4.5 + ……. + k(k + 1) (k + 2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\)

We show that the statement is true for n = k + 1

(i.e) We show that S(k + 1) = \(\frac{(k+1)(k+2)(k+3)(k+4)}{4}\)

We observe that

S(k + 1) = 1.2.3 + 2.3.4 + …… + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)

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

= \(\frac{k(k+1)(k+2)(k+3)}{4}\) + (k + 1)(k + 2)(k + 3)

= (k + 1)(k + 2)(k + 3) \(\left(\frac{k}{4}+1\right)\)

= \(\frac{(k+1)(k+2)(k+3)(k+4)}{4}\)

∴ The statement holds for n = k + 1

∴ By the principle of mathematical induction,

p(n) is true for all n ∈ N

(i.e.,) 1.2.3 + 2.3.4 + 3.4.5 + ……. + (n)(n + 1)(n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)

Question 14.

\(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}\) + …. up to n terms = \(\frac{n}{24}\) [2n^{2} + 9n + 13]

Solution:

The nth term of the given series is \(\frac{1^{3}+2^{3}+3^{3}+\ldots .+n^{3}}{1+3+5+\ldots+(2 n-1)}\)

Let p(n) be the statement :

and let S(n) be the sum on the L.H.S.

∵ S(1) = \(\frac{1^{3}}{1}=\frac{1}{24}(2+9+13)=1=\frac{1^{3}}{1}\)

∴ p(1) is true

Assume that p(k) is true

(i.e.,) S(k) = \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\ldots+\frac{1^{3}+2^{3}+\ldots \pm k^{3}}{1+3+\ldots+(2 k-1)}\) = \(\frac{k}{24}\) [2k^{2} + 9k + 13]

We show that p(k + 1) is true

(i.e,) we show that

∴ The statement holds for n = k + 1

∴ By the principle of mathematical induction,

p(n) is true for all n

(i.e.,) \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\ldots+\frac{1^{3}+2^{3}+\ldots \ldots+n^{3}}{1+3+\ldots+(2 n-1)}\) = \(\frac{n}{24}\) [2n^{2} + 9n + 13]

Question 15.

1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + ……. up to n terms = \(\frac{n(n+1)^{2}(n+2)}{12}\)

Solution:

The nth term of the given series is (1^{2} + 2^{2} + 3^{2} + …… + n^{2})

Let p(n) be the statement:

1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + ……. + (1^{2} + 2^{2} + …… + n^{2}) = \(\frac{\mathrm{n}(\mathrm{n}+1)^{2}(\mathrm{n}+2)}{12}\)

and the sum on the LH.S. is denoted by S(n).

Since S(1) = 1^{2} = \(\frac{1(1+1)^{2}(1+2)}{12}\) = 1 = 1^{2}

∴ p(1) is true.

Assume that the statement is true for n = k

(i.e.,) S(k) = 1^{2} + (1^{2} + 2^{2}) + ……. + (1^{2} + 2^{2} + ……. + k^{2})

= \(\frac{k(k+1)^{2}(k+2)}{12}\)

We show that S(k + 1) = \(\frac{(k+1)(k+2)^{2}(k+3)}{12}\)

We observe that

∴ The statement holds for n = k + 1.

∴ By the principle of mathematical induction,

p(n) is true for all n ∈ N.

(i.e.,) 1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + …….. (1^{2} + 2^{2} + ………. + n^{2}) = \(\frac{n(n+1)^{2}(n+2)}{12}\)