Students must practice these TS Intermediate Maths 1A Solutions Chapter 2 Mathematical Induction Ex 2(a) to find a better approach to solving the problems.

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

Using Mathematical Induction prove each of the following statement for all n ∈ N.

Question 1.

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

Answer:

Let S(n) be the given statement

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

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

⇒ 1 = 1; the statement is true for n = 1

Suppose the statement is true for n = k then

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

We have to prove that the statement is true for n = k + 1 also then

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

∴ The statement is true for n = k + 1 also.

∴ By the principle of Finite Mathematical Induction S(n) is true for all n ∈ N.

i.e., 1^{2} + 2^{2} + 3^{2} + ……….. + n^{2} = \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\), ∀ n ∈ N

Question 2.

2.3 + 3.4 + 4.5 + ………………. upto n terms = \(\frac{n\left(n^2+6 n+11\right)}{3}\) (March 13, May 06)

Answer:

Let S(n) be the statement.

The n^{th} term of 2.3 + 3.4 + 4.5 + …………… is (n + 1) (n + 2)

∴ 2 . 3 + 3 . 4 + 4 . 5 + …………….. + (n + 1) (n + 2)

= \(\frac{n\left(n^2+6 n+11\right)}{3}\)

Now S(1) = 2 . 3 = \(\frac{1\left(1^2+6+11\right)}{3}\) = 6

∴ The statement is true for n = 1.

Suppose that the statement is true for n = k, then 2.3 + 3.4 + 4.5 + …………….. + (k + 1) (k + 2)

= \(\frac{k\left(k^2+6 k+11\right)}{3}\)

Adding (k + 1) th term of L.H.S both sides

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

∴ S(k + 1) = \(\frac{1}{3}\) (k + 1) [k^{2} + 2k + 1 + 6 (k + 1) + 11]

= \(\frac{1}{3}\) (k + 1) [(k + 1)^{2} + 6(k + 1) + 11]

∴ The statement is true for n = k + 1

So by the principle of Finite Mathematical Induction S(n) is true ∀ n ∈ N

∴ 2 . 3 + 3 . 4 + 4 . 5 + ……………. + (n + 1) (n + 2) = \(\frac{n\left(n^2+6 n+11\right)}{3}\)

Question 3.

\(\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}\) + ……………. + \(\frac{1}{(2 n-1)(2 n+1)}\) = \(\frac{n}{2 n+1}\) (May 2014)

Answer:

Let S_{n} be the statement

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

Then S(1) = \(\frac{1}{1 \cdot 3}=\frac{1}{2(1)+1}=\frac{1}{3}\)

∴ S(1) is true.

Suppose the statement is true for n = k, then

S(K) = \(\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots .+\frac{1}{(2 k-1)(2 k+1)}\) = \(\frac{\mathbf{k}}{2 \mathbf{k}+1}\)

We have to show that the statement is true for n = k + 1 also,

The statement S(n) is true for n = k + 1

∴ By the principle of Mathematical Induction S(n) is true for all n ∈ N.

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

Question 4.

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

Answer:

4, 8, 12, are in A.P. and nth term of A.P. is = a + (n – 1) d

= 4 + (n – 1) 4 = 4n

Let S(n) be the statement

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

Let n = 1, then

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

∴ The statement is true for n = 1 also.

Suppose the statement is true for n = k then

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

We have to prove that the result is true for n = k + 1 also. Adding (k + 1) th term

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

4^{3} + 8^{3} + 12^{3} + ……………….. + (4k)^{3} + (4k + 4)^{3}

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

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

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

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

Hence the result is true for n = k + 1.

∴ By the principle of Mathematical Induction S(n) is true ∀ n ∈ N.

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

Question 5.

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

Answer:

Let S(n) be the statement

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

Now S(1) is a = \(\frac{1}{2}\) [2a + 0 (d)] = a

∴ S(1) is true.

Assume that the statement is true for n = k

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

= \(\frac{k}{2}\) [2a + (k – 1) d]

We have to prove that the statement is true for n = k + 1 also.

Adding a + kd both sides

a + (a + d) + ……………. + [a + (k – 1) d] + [a + kd]

∴ The statement is true for n = k + 1 also

∴ By the principle of Mathematical Induction.

S(n) is true for all n ∈ N

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

Question 6.

a + ar + ar^{2} + ………………… + n terms = \(\frac{a\left(r^n-1\right)}{r-1}\), r ≠ 1 (March 2011)

Answer:

Let S(n) be the statement

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

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

∴ The result is true for n = 1

Suppose the statement is true for n = k then

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

We have to prove that the result is true for n = k + 1 also.

Adding ar^{k} both sides

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

∴ The statement is true for n = k + 1 also

∴ By the principle of Mathematical Induction p(n) is true for all n ∈ N

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

Question 7.

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

Answer:

Let S(n) be the statement

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

Since S(1) = 2,

S(1) is true.

Suppose the statement is true for n k then

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

We have to show that S(n) is true for n = k + 1 also.

Adding (k + 1)th term 5 (k + 1) – 3 = 5k + 2 both sides

[2 + 7 + 12 + ……. + 5k – 3)] + (5k + 2)

∴ S(n) is true for n = k + 1 also

∴ By the principal of Mathematical Induction

S(n) is true ∀ n ∈ N

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

Question 8.

\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots \ldots\left(1+\frac{2 n+1}{n^2}\right)\) = (n + 1)^{2} (March 2015-A.P)

Answer:

Let S_{n} be the statement

∴ S(n) is true for n = 1

Suppose S_{n} is true for n = k then

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

= (k + 1)^{2} ………………. (1)

We have to prove that the statement is true for n = k + 1 also

(k + 1) th term is

= k^{2} + 4k + 4

= (k + 2)^{2}

∴ S(n) is true for n = k + 1 also

∴ By the principal of Mathematical Induction S(n) is true for ∀ n ∈ N

Question 9.

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

Answer:

Let S(n) be the statement

When n = 1, then 9 < 16

∴ S(n) is true for n = 1

Suppose S(n) is true for n = k

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

We have to prove that the result is true for n = k + 1

i.e., 2(k + 1) + 7 < (k + 4)^{2}

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

= k^{2} + 6k + 9 + 2

= k^{2} + 6k + 11 < (k^{2} + 6k + 11) + (2k + 5)

= k^{2} + 8k + 16

= (k + 4)^{2}

∴ S(n) is true for n = k + 1 also

By principle of Mathematical Induction

S(n) is true ∀ n ∈ N

Question 10.

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

Answer:

Let S(n) be the statement

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

∴ S(n) is true for n = 1

Assume S(n) to be true for n = k then

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

We have to prove that the result is true for n = k + 1 also.

∴ S(n) is true for n = k + 1 also

∴ By principle of Mathematical Induction.

S(n) is true ∀ n ∈ N

Question 11.

4^{n} – 3n – 1 is divisible by 9

Answer:

Let S(n) be the statement,

4^{n} – 3n – 1 is divisible by 9

For n = 1, 4 – 3 – 1 = 0 is divisible by 9

∴ Statement S(n) is true for n = 1.

Suppose the statement S(n) is true for n = k

Then 4^{k} – 3k – 1 is divisible by 9.

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

We have to show that statement is true for n = k + 1 also.

From (1), 4^{k} = 9t + 3k + 1

∴ 4^{k + 1} – 3 (k + 1) – 1 = 4 . 4^{k} – 3 (k + 1) – 1

= 4 (9t + 3k + 1) – 3k – 3 – 1

= 4 (9t) + 9k

= 9 (4t + k) divisible by 9

(∵ 4t + k is an integer)

Hence, S(n) is true for n = k + 1 also.

∴ 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.

S(n) is true for all n e K

∴ 4n – 3n – 1 is divisible by 9

Question 12.

3.5^{2n + 1} + 2^{3n + 1} is divisible by 17 (May 2012, 2008)

Answer:

Let S_{n} be the statement

3.5^{2n + 1} + 2^{3n + 1} is divisible by 17

S(1) is 3 . 5^{2(1) + 1} + 2^{3 (1) + 1}

= 3 . 5^{3} + 2^{4} = 3 (125) + 16

= 375 + 16 = 391 is divisible by 17

Hence, S(n) is true for n = 1.

Suppose that the statement is true for n = k, then

3.5^{2k + 1} + 2^{3k + 1} is divisible by 17 and

3.52^{k + 1} + 2^{3k + 1} = 17t for t ∈ N

then we have to show that the result is true for n = k + 1 also

Consider 3.5^{2(k + 1)} + 1 + 2^{3(k + 1)} + 1

= 3.5^{2k + 1} . 5^{2} + 2^{3(k + 1)} . 2

= (17t – 2^{3k + 1}) 5^{2} + 2^{3k + 3} . 2

= 17t (25) – 2^{3k} (50) + 2^{3k} (16)

= 17 t (25) + 2^{3k} (16 – 50)

= 17 t (25) – 34 2^{3k}

= 17 [25t – 2^{3k + 1}

25t – 2^{3k + 1} is an integer.

∴ 3.5^{2(k + 1) + 1} + 2^{3 (k + 1) + 1} is divisible by 17.

∴ The statement S(n) is true for n = k + 1 also.

∴ By the principle of Mathematical induction S(n) is true for ∀n ∈ N

∴ 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 = \(\frac{n(n+1)(n+2)(n+3)}{4}\). (March 2015-T.S) (Mar. 08)

Answer:

The n^{th} term of the given series is n (n + 1) (n + 2) and let S_{n} be the statement.

1.2.3 + 2.3.4 + 3.4.5 + ……………. + n (n + 1) (n + 2)

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

For n = 1

S(1) = 1.2.3 = 6

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

= \(\frac{2(3)(4)}{4}\) = 6

∴ S(n) is true for n = 1

Let S(n) is true for n = k then

1.2.3 + 2.3.4 + 3.4.5 + ……………. + k (k + 1) (k + 2)

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

We have to prove that the result is true for n = k + 1 also.

Adding (k + 1) th term, (k + 1) (k + 2) (k + 3) both sides we get

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

∴ S(n) is true for n = k + 1 also.

Hence by the principle of Mathematical Induction.

S(n) is true for ∀n ∈ N

∴ 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.

\(\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}\) + ……………… upto n terms = \(\frac{n}{24}\) (2n^{2} + 9n + 13). (Mar. 14, 04, 05)

Answer:

The n^{th} term of the given series is

Question 15.

1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + ……………. upto n terms = \(\frac{n(n+1)^2(n+2)}{12}\). (March 2012)

Answer:

Let S(n) be the statement and

nth term of series is 1^{2} + 2^{2} + 3^{2} + ……………… + n^{2}

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

∴ S(n) is true for n = 1.

Suppose S(n) is true for n = k then

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

Now we have to prove that S(k + 1) is true.

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

∴ The statement S(n) hold for n = k + 1 also

∴ By the principle of Mathematical Induction S(n)is true ∀n ∈ N.

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