## MP Board Class 11th Maths Important Questions Chapter 4 Principle of Mathematical Induction

### Principle of Mathematical Induction Important Questions

**Principle of Mathematical Induction Long Answer Type Questions**

Question 1.

For n > 1, prove that with the help of principle of mathematical induction :

1^{2} + 2^{2} + 3^{2} + 4^{2} + …………. + n^{2} = \(\frac { n(n + 1)(2n + 1) }{ 6 }\). (NCERT)

Solution:

∴ p(n) is true for n = k + 1.

Hence, p(n) is true for all values of n ∈ N.

Question 2.

For n ≥ 1 prove that :

\(\frac { 1 }{ 1.2 }\) + \(\frac { 1 }{ 2.3 }\) + \(\frac { 1 }{ 3.4 }\) + …………. + \(\frac { 1 }{ n(n + 1) }\) = \(\frac { n }{ n + 1 }\)

Solution:

Let

∴ p(n) is true for n = k + 1.

Hence, p(n) is true for all values of n ∈ N.

Instruction:

All questions from 3 to 7 when n∈N prove the statement by using principle of mathematical induction :

Question 3.

1^{3} + 2^{3} + 3^{3} + …………. + n^{3} = \(\frac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 4 }\) (NCERT)

Solution:

∴ Given function is true for n = m + 1.

Hence, p(n) is true for all values of n ∈ N.

Question 4.

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

Solution:

For n = 1, L.H.S = 1

and R.H.S. = \(\frac { 2 x 1 }{ 1 + 1 }\) = 1

The given function is true for n = 1.

Let the function be true for n = m

∴ The Given function is true for n = m + 1.

Hence, p(n) is true for all values of n ∈ N.

Question 5.

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

Solution:

Let p(n) = 1.2 + 2.3 + 3.4 + ………….. + n(n + 1) = \(\frac { n(n + 1)(n + 2) }{ 3 }\)

If n = 1, L.H.S. = 1.2 = 2

R.H.S = \(\frac { n(n + 1)(n + 2) }{ 3 }\) = \(\frac { 1.2.3 }{ 3 }\) = 2

∴ p(n) is true for n = 1.

Let p(n) be true for n = k

∴ p(n) is true for n = k + 1.

Hence, p(n) is true for all values of n ∈ N.

Question 6.

\(\frac { 1 }{ 2.5}\) + \(\frac { 1 }{ 5.8 }\) + \(\frac { 1 }{ 8.11}\) + …………. + \(\frac { 1 }{ (3n – 1)(3n + 2) }\) = \(\frac { n }{ 6n + 4 }\)

Solution:

Let p(n) = \(\frac { 1 }{ 2.5}\) + \(\frac { 1 }{ 5.8 }\) + \(\frac { 1 }{ 8.11}\) + …………. + \(\frac { 1 }{ (3n – 1)(3n + 2) }\) = \(\frac { n }{ 6n + 4 }\)

∴ For n = 1, L.H.S = \(\frac { 1 }{ 2.5}\) = \(\frac { 1 }{ 10}\)

and R.H.S = \(\frac { 1 }{ 6.1 + 4}\) = \(\frac { 1 }{ 10}\)

∴ p(n) is true for n = 1.

Let p(n) be true for n = k

∴ p(n) is true for n = k + 1.

Hence, p(n) is true for all values of n ∈ N.

Question 7.

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

Solution:

For n = 1, L.H.S = 1.2.3 = 6

And R.H.S. = \(\frac { 1(1 + 1)(1 + 2)(1 + 3) }{ 4 }\) = \(\frac { 2.3.4 }{ 4 }\) = 6

∴ Given function is true for n = 1.

Let the function be true for n = k

∴ The given function is true for n = k + 1. Hence the given function is true for all values of n ∈ N.

Question 8.

Prove that (41)^{n} – (14)^{n} is a multiple of 27 with the help of principle of mathematical induction. (NCERT)

Solution:

Let P(n) = (41)^{n} – (14)^{n}

For n = 1,

P(1) = (41)^{1} – (14)^{1}= 27

Which is multiple of 27.

Let P(ri) be true for n = k.

Then, (41)^{k} – (14)^{k} = 27m

Put n = k +1 in P(n)

(41)^{k+1} – (14)^{k+1} = (41)^{k}.41 – (14)^{k}. 14

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

= 41[(41)^{k} – (14)^{k}]+(14)^{k} [41 – 14]

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

Which is multiple of 27.

If the given function P(n) is true for n = k then P(n) will be true for U n = k +1 also.

Hence the given function is true for all values of n ∈ N.