ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

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ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Using the principle of mathematical induction prove that (1 to 22) for all n ∈ N :

Question 1.
2 + 4 + 6 + …………… + 2n = n² + n.
Solution:
Let P (n) be the statement :
2 + 4 + 6+ ………………….. + 2n = n² + n
Now P (1) means, 2 = 1² + 1 = 2, which is true
= P(1) is true.
Let P (m) be true
i.e. 2 + 4 + 6 + ……………….. + 2m = m² + m
For P (m + 1) ;
2 + 4 + 6 …………………. + 2 (m + 1) = 2 + 4 + 6 + ………….. + 2m + 2m + 2
= m² + m + 2m + 2 [using (1)]
= m² + 3m + 2
= (m + 1) (m + 2)
= (m + 1) [m + 1 + 1]
= (m + 1)² + (m + 1)
Thus, P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 2.
1 + 4 + 7 + ……………………. + (3n – 2) = \(\frac{1}{2}\) n (3n – 1)
Solution:
Let P (n) be the statement :
1 + 4 + 7 + ……………………. + (3n – 2) = \(\frac{1}{2}\) n (3n – 1)
Now P (1) means,
1 = \(\frac{1}{2}\) × (3 – 1)
= \(\frac{1}{2}\) × 2 = 1, which is true
⇒ P(1) is true.
Let P (m) is true i.e. 1 + 4 + 7 + ……………………….. + 3m – 2 = \(\frac{1}{2}\) m (3m – 1)
For P (m + 1) :
1 + 4 + 7 …………… + 3m – 2 = \(\frac{1}{2}\) m (3m – 1)
= \(\frac{1}{2}\) m (3m – 1) + 3m + 1
= \(\frac{1}{2}\) [3m² – m + 6m + 2]
= \(\frac{1}{2}\) [3m² + 5m + 2]
= \(\frac{1}{2}\) (m + 1) (3m + 2)
= \(\frac{1}{2}\) (m + 1) [3 (m + 1) – 1]
Thus P(m + 1) is true.
Hence, by mathematical induction, P (n) is true for all n ∈ N.

Question 3.
3x+ 6x+ 9x + …………………………. to n terms = \(\frac{3}{2}\) n (n + 1) x.
Solution:
Let P (n) be the statemen t:
3x + 6x + 9x …………………….. + n terms = \(\frac{3}{2}\) n (n + 1) x
Now P (1) means,
3x = \(\frac{3}{2}\) 1 (1 + 1)x
= \(\frac{3}{2}\) × 2x = 3x, which is true
∴ P (1) is true.
Let P (m) is true i.e. 3x + 6x + 9x + ………………….. + m terms = \(\frac{3}{2}\) m (m + 1) x ……………….(1)
For P (m + 1) :
3x + 6x + 9x + ……………… + (m + 1) terms = 3x + 6x + 9x + ………….. + 3mx + 3 (m + 1) x
= \(\frac{3}{2}\) m (m + 1) x + 3 (m + 1) x [using (1)]
= \(\frac{3(m+1) x}{2}\) [m + 2]
= \(\frac{3}{2}\) (m + 1) (m + 1 + 1) x
⇒ P(m + 1) is true.
Thus by mathematical induction, P (n) is true for all n ∈ N.

Question 4.
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^n}=1-\frac{1}{2^n}\).
Solution:
Let P (n) be the statement:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^n}=1-\frac{1}{2^n}\)
Now P (1) means,
\(\frac{1}{2}=1-\frac{1}{2^1}=1-\frac{1}{2}=\frac{1}{2}\) which is true
⇒ P (1) is true.
Let P (m) is true.
i.e. \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^m}=1-\frac{1}{2^m}\) ………………..(1)
Now P (m + 1) ;
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^m}+\frac{1}{2^{m+1}}\)
= \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8} \ldots .+\frac{1}{2^m}\right)+\frac{1}{2^{m+1}}\)
= 1 – \(\frac{1}{2^m}+\frac{1}{2^{m+1}}\) [using (1)]
= 1 – \(\frac{1}{2^m}\left[1-\frac{1}{2}\right]\)
= 1 – \(\frac{1}{2^m} \times \frac{1}{2}\)
= 1 \(\frac{1}{2^{m+1}}\)
⇒ P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 5.
1² + 3² + 5² + ……………… to n terms = \(\frac{n\left(4 n^2-1\right)}{3}\)
Solution:
Let p (n) be the statement :
1² + 3² + 5² + ……………… to n terms = \(\frac{n\left(4 n^2-1\right)}{3}\)
Now P (1) means 1² = 1, which is true
∴ P (1) is true.
Let P (m) be true.

Question 6.
1³ + 2³ + ……………… + n³ = \(\left(\frac{n(n+1)}{2}\right)^2\)
Solution:
Hence P(n) be the statement :
1³ + 2³ + ……………… + n³ = \(\left(\frac{n(n+1)}{2}\right)^2\)
For n = 1 ;
P (1) means n³ = \(\left(\frac{1(1+1)}{2}\right)^2\) = 1, which is true.
∴ P(1) is true
Let P (n) be true for n = m
Thus,
1³ + 2³ + ……………… + m³ = \(\left(\frac{m(m+1)}{2}\right)^2\) ………………..(1)
For P (m + 1) :
1³ + 2³ + …………………. + m³ + (m + 1)³ = {1³ + 2³ + ………………….. + m³} + (m + 1)³
= \(\left[\frac{m(m+1)}{2}\right]^2\) + (m + 1)³ [using (1))
= \(\frac{(m+1)^2}{4}\) [m² + 4m + 4]
= \(\frac{(m+1)^2(m+2)^2}{4}\)
= \(\left[\frac{(m+1)(\overline{m+1}+1)}{2}\right]^2\)
Thus P(n) is true for n = m + 1.
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 7.
3 . 6 + 6 . 9 + 9. 12 + ……………………… + 3n (3n + 3) = 3n (n + 1) (n + 2).
Solution:
Let P (n) be the statement :
3 . 6 + 6 . 9 + 9. 12 + ……………………… + 3n (3n + 3) = 3n (n + 1) (n + 2)
Now P (1) means,
3.6 = 3 × 1 (1 + 1) (1 + 2)
⇒ 18 = 3 × 2 × 3 = 18, which is true
∴ P (1) is true.
Let P (m) is true
i.e. 3 . 6 + 6 . 9 + 9 . 12 + …………………….. + 3m (3m + 3) = 3m (m + 1) (m + 2)
For P (m + 1) :
3 . 6 + 6 . 9 + 9 . 12 + …………………….. + 3m (3m + 3) + 3m (m + 1) (m + 2) = 3m (m + 1) (m + 2) + 3 (m + 1) (3m + 6) [using (1)]
= 3 (m + 1) [m (m + 2) + 3m + 6]
= 3 (m +1) [m2 + 5m + 6]
= 3 (m + 1) (m + 2) (m + 3)
= 3 (m + 1) (m + 1 + 1) (m + 1 + 2)
⇒ P (m + 1)is true.
Hence by induction P(n) is truc for all n ∈ N.

Question 8.
1 . 2 . 3 + 2 . 3 . 4 + …………………… + n (n + 1) (n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
Solution:
Let P (n) be the statement :
1 . 2 . 3 + 2 . 3 . 4 + …………………… + n (n + 1) (n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
For n = 1 ;
P (1) = 1 . 2 . 3 = 6
= \(\frac{1(1+1)(1+2)(1+3)}{4}\)
= \(\frac{24}{4}\) = 6
Thus P (1) is true.
Let us assume that P (n) is true for n = m.
∴ 1 . 2 . 3 + 2 . 3 . 4 + …………………… + m (m + 1) (m + 2) = \(\frac{m(m+1)(m+2)(m+3)}{4}\) …………………..(1)
For P (m + 1) ;
1 . 2 . 3 + 2 . 3 . 4 + …………………… + m (m + 1) (m + 2) + (m + 1) (m + 2) (m + 3) = \(\frac{m(m+1)(m+2)(m+3)}{4}\) + (m + 1) (m + 2) (m + 3)
= \(\frac{(m+1)(m+2)(m+3)}{4}\) [m + 4]
= \(\frac{(m+1)(\overline{m+1}+1)(\overline{m+1}+2)(\overline{m+1}+3)}{4}\)
∴ P (n) is true forn=m+I
Hence by mathematical induction result is true V n E N.

Question 9.
1 . 2 + 2 . 2² + 3 . 2³ + ………………. + n . 2ⁿ = (n – 1) . 2ⁿ⁺¹ + 2.
Solution:
Let P (n) be the statement:
1 . 2 + 2 . 2² + 3 . 2³ + ………………… + n . 2ⁿ (n – 1) 2ⁿ⁺¹ + 2
Here P (1) :
1 . 2 = 2 = (1 – 1) 2¹⁺¹ + 2 = 2, which is true
∴ P (1) is true.
Let P(n) is true for n = m
i.e. 1 . 2 + 2 . 2² + ………………… + m . 2^m = (m – 1) 2^m+1 + 2 ………………..(1)
For P (m + 1) :
1 . 2 + 2 . 2² + ………………….. + m . 2^m + (m + 1) 2^{m + 1}
= (m – 1) 2^{m + 1} + (m + 1) 2^m+1 + 2 [using (1)]
= 2^m+1 (m – 1 + m ÷ I) + 2
= m . 2^{m + 2} + 2
∴ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

Question 10.
\(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{n}\right)\) = n + 1
Solution:
Let P (n) be the statement :
\(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{n}\right)\) = n + 1
Now P (1) means,
1 + \(\frac{1}{1}\) = 1 + 1, which is true
∴ P (1) is true.
Let P (m) is true.
i.e. \(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right) \ldots\left(1+\frac{1}{m}\right)\) = m + 1 …………………(1)
For P (m + 1) ;
\(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right) \ldots\left(1+\frac{1}{m}\right)\left(1+\frac{1}{m+1}\right)=(m+1)\left[1+\frac{1}{m+1}\right]\)
= m + 1 + 1
⇒ P (m + 1) is true.
Thus by mathematical induction, result P (n) is true for all n ∈ N.

Question 11.
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \cdots\left(1-\frac{1}{n+1}\right)=\frac{1}{n+1}\).
Solution:
Let P (n) be the statement :
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \cdots\left(1-\frac{1}{n+1}\right)=\frac{1}{n+1}\)
Now P (1) means,
\(1-\frac{1}{2}=\frac{1}{1+1}\)
i.e. \(\frac{1}{2}=\frac{1}{2}\), which is true.
⇒ P (1) is true.
Let P (m) is true.
i.e. \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots\left(1-\frac{1}{m+1}\right)=\frac{1}{m+1}\) ………………….(1)
For P (m + 1) ;
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots\left(1-\frac{1}{m+1}\right)\left(1-\frac{1}{m+2}\right)\)
= \(\frac{1}{m+1}\left(1-\frac{1}{m+2}\right)\)
= \(\frac{1}{m+1}\left[\frac{m+2-1}{m+2}\right]\)
= \(\frac{1}{m+2}\)
= \(\frac{1}{m+1+1}\)
⇒ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

Question 12.
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\) + ……………………. to n terms = \(\frac{n}{n+1}\)
Solution:
Let P(n) be the statement :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\) + ……………………. to n terms = \(\frac{n}{n+1}\)
i.e. \(\frac{1}{1.2}+\frac{1}{2.3}+\ldots .+\frac{1}{n(n+1)}=\frac{n}{n+1}\)
[∵ nth term of (1, 2, 3, …………………n terms) (2, 3, 4, ………………. n terms) = [1 + (n – 1) 1] [2 + (n – 1) . 1] = n (n + 1)]
Now P (1) means,
\(\frac{1}{1.2}=\frac{1}{1+1}\)
⇒ \(\frac{1}{2}=\frac{1}{2}\), which is true.
⇒ P (1) is true.
Let P (m) be true.

Question 13.
\(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\ldots+\frac{1}{3 n(3 n+3)}=\frac{n}{9(n+1)}\)
Solution:
\(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\ldots+\frac{1}{3 n(3 n+3)}=\frac{n}{9(n+1)}\)

Question 14.
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\) + …………….. to n terms = \(\frac{n}{2 n+1}\).
Solution:
nth term gives series = nth term of (1, 3, 5, ………………… n terms) (nth term of 3 . 5, 7, ……………..)
= [1 + (n – 1) 2] + [3 + (n – 1) 2]
= (2n – 1) (2n + 1)
Let P (n) be the statement :

Question 15.
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}\).
Solution:
Let P (n) be the statement :
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}\)
Now P (1) means,
\(\frac{1}{1.4}=\frac{1}{3+1}\)

Question 16.
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}\)
Solution:
Let P (n) be the statement :
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}\)
Now P (1) means,
\(\frac{1}{3.5}=\frac{1}{3(2+3)}\)
⇒ \(\frac{1}{15}=\frac{1}{15}\), which is true
⇒ P (1) is true.
Let P (m) be true.
i.e. \(\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 m+1)(2 m+3)}=\frac{m}{3(2 m+3)}\) …………………..(1)

Question 17.
2²ⁿ – 1 is divisible by 3.
Solution:
Let P (n) be the statement :
2²ⁿ – 1 is divisible by 3
Now p (1) means, 2^{2 × 1} – 1 is divisible by 3
⇒ 3 is divisible by 3, which is true
⇒ P (1) is true.
Let P (m) is true
i.e. 2^2m – 1 be divisible by 3
⇒ 2^2m – 1 = 3k for some integer k ……………….(1)
For P (m + 1) :
2^{2 (m + 1)} – 1 = 2^{2m + 2} – 1
= 2^2m . 4 – 1
= (3k + 1) 4 – 1 [using (1)]
= 12k + 3
= 3 (4k + 1)
[since k ∈ 1 ∴ 4k + 1 ∈ I]
which is divisible by 3.
∴ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

Question 18.
2³ⁿ – 1 isdivisible by 7.
Solution:
Let P (n) be the statement:
2³ⁿ – 1 is divisible by 7.
Now P (1) means,
2^{3 × 1} – 1 is divisible by 7.
i.e. (8 – 1) i.e. 7 is divisible by 7.
⇒ P (1) is true.
Let P (m) be true
i.e. 2^3m – 1 is divisible by 7.
⇒ 2^3m – 1 = 7k for some integer k
⇒ 2^3m = 7k + 1 …………………..(1)
For P (m + 1) ;
2^{3 (m + 1)} – 1 = 2^{3m + 3} – 1
= 2^3m – 8 – 1
= 8 (7k + 1) – 1 [using (1)]
= 56k + 7
= 7 (8k + 1).
which is divisible by 8. [∵ k ∈ I
⇒ 8k + 1 ∈ I]
P (m + 1) is true
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 19.
3²ⁿ when divided by 8, leaves the remainder 1.
Solution:
We want to prove that,
3²ⁿ when divided by 8, leaves the remainder I. i.e. 3²ⁿ – 1 divisible by 8.
Let P (n) be the statement:
3²ⁿ – 1 is divisible by 8.
Now P (1) means, 3^{2 × 1} – 1 is divisible by 8
⇒ (9 – 1) = 8 is divisible by 8
⇒ P(1) is true.
Let P(m) be true i.e. 3^2m – 1 is divisible by 8
⇒ 3^2m – 1 = 8k for some integer k …………………(1)
For P (m + ) :
3^{2 (m + 1) – 1} = 3^2m . 3² – 1
= 9 (8k + 1) – 1 [using (1)]
= 72k + 9 – 1
= 8 [9k + 1]
= 8k’ [where k’ = 9k + 1 ∈ I]
which is divisible by 8
⇒ P (m + 1) is true.
Thus by induction, P(n) is true for all n ∈ N.

Question 20.
41ⁿ – 14ⁿ is divisible by 27.
Solution:
Let P (n) be the statement :
41ⁿ – 14ⁿ is divisible by 27.
P(1) means, 41¹ – 14¹ = 27 which is divisible by 27.
∴ P(1) is true.
Let us assume that P (n) is true for n = m
i.e. 41^m – 14^m is divisible by 27.
⇒ 41^m – 14^m = 27k where k ∈ N …………………(1)
For P (m + 1) ;
41^{m + 1} – 14^{m + 1}
= 41 (27k + 14^m) – 14^m+1
= 41 × 27k + 14^m (41 – 14)
= 27 (41k + 14^m)
= 27k’
where k’ = 41k + 14^m ∈ N
which is divisible by 27.
Thus, P (m + 1) is true.
Hence by M.I., P (n) is true for all n ∈ N.

Question 21.
7²ⁿ + 2^{3n – 3} . 3^{n – 1} is divisible by 25.
Solution:
Let P (n) be the statement :
7²ⁿ + 2^{3n – 3} . 3^{n – 1} is divisible by 25.
Here P (1) means,
7² + 2^{3 – 3} 3^{1 – 1} = 49 + 1 = 50 which is divisible by 25.
∴ P(1) is true.
Let us assume that P (n) is true for n = m
i.e. 7^2m + 2^{3m – 3} . 3^{m – 1} is divisible by 25.
⇒ 7^2m + 2^{3m – 3} . 3^{m – 1} = 25k ………………..(1)
P (m + 1) : 7^{2 (m + 1)} + 2^{3 (m + 1) – 3} . 3^{m + 1 – 1}
= 7^{2 (m + 1)} + 2^3m . 3^m
= 7² . (25k – 23m – 3 . 3m – 1) + 2^3m 3^m [using (1)]
= 49 × 25k – 2^{3m – 3} 3^{m – 1} – 1(49 – 8 × 3)
= 25 (49k – 2^{3m – 3} 3^{m – 1}) = 25k’
where k’ = 49k – 2^{3m – 3} 3^{m – 1} ∈ N
which is clearly divisible by 25
Thus P (m + 1) is true.
Hence, by mathematical induction, P (n) is true ∀ n ∈ N.

Question 22.
7ⁿ – 3ⁿ is divisible by 4.
Solution:
Let P(n) be the statement :
7ⁿ – 3ⁿ is divisible by 4.
Now P (1) means,
7¹ – 3¹ is divisible by 4
i.e. 4 is divisible by 4 which is true
⇒ P (1) is true.
Let P (m) be true
Le. 7^m – 3^m is divisible by 4
⇒ 7^m – 3^m = 4k for some integer k ………………..(2)
For P (m + 1) ;
7^{m + 1} – 3^{m + 1} = 7^m . 7 – 3^{m + 1}
= 7 (3^m + 4k) – 3^{m + 1} [using (1)]
= 7 . 3^m + 28k – 3^m . 3
= 3^m (7 – 3) + 28k
= 4 [3^m + 7k] = 4k’
which is divisible by 4
[where k’ = 3^m + 7k ∈ I, where k ∈ I]
⇒ P(m + 1) is true.
Hence by induction, P (n) is true for all n ∈ N.

Question 23.
4ⁿ + 15n – 1 is divisible by 9.
Solution:
Let P (n) be the statement:
4ⁿ + 15n – 1 is divisible by 9
Now P (1) means,
4¹ + 15 – 1 = 18 is divisible by 9,
which is true
⇒ P(1) is true.
Let P (m) be true
i.e. 4^m + 15m – 1 is divisible by 9
⇒ 4^m + 15m – 1 = 9k for some integer k
⇒ 4^m + 15m = 9k + 1 …………… (1)
For P (m + 1) :
4^{m + 1} + 15 (m + 1) – 1 = 4 . 4^m + 15m + 14
= 4 (- 15m + 9k + 1) + 15m + 14 [using (1)]
= – 45m + 36k + 18
= 9 (- 5m + 4k + 2) = 9k’
where k’ = – 5m + 4k + 2 ∈ I
which is divisible by 9.
⇒ P(m + 1) is true.
Hence by mathematical induction, P (n)is true for all n ∈ N.

Question 24.
3^{2n + 2} – 8n – 9 is a multiple of 64.
Solution:
Let P (n) be the statement:
3^{2n + 2} – 8n – 9 is a multiple of 64.
Now P (1) means,
3^{2 + 2} – 8 × 1 – 9 is multiple of 64.
i.e. 81 – 8 – 9 = 64 is a multiple of 64,
which is true
⇒ P(1) is true.
Let P (m) is true
i.e. 3^{2m + 2} – 8m – 9 is a multiple of 64
⇒ 3^{2m + 2} – 8m – 9 = 64k for some integer k
⇒ 3^{2m + 2} = 8m + 9 + 64k ……………(1)
For P (m + 1) ;
3^{2m + 4} – 8 (m + 1) – 9
= 3^{2m + 2} . 3² – 8 (m + 1) – 9
= 9 (8m + 9 + 64k) – 8m – 17 [using(1)]
= 64m + 64k + 64
= 64 (m + k + 1)
= 64k’ where k’ = m + k + 1 ∈ I
which is divisible by 64.
⇒ P(m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 25.
n (n + 1) (n + 5) is a multiple of 3.
Solution:
Let P (n) be the statement :
n (n + 1) (n + 5) is a multiple of 3.
Now P (1) means,
1 (1 + 1) (1 + 5) = 2 × 6 = 12 is a multiple of 3.
which is true
∴ P (1) is true.
Let P (m) be true i.e. m (m + 1) (m + 5) is a multiple of 3
⇒ m (m + 1) (m + 5) = 3k for some integer k
⇒ m (m² + 8m + 5) = 3k
For P (m + 1) ;
(m + 1) (m + 2) (m + 6) = (m + 1) [m² + 8m + 12]
= m (m² + 8m + 12) + m² + 8m + 12
= m (m² + 6m + 5) + m (2m + 7) + m² + 8m + 12
= 3k + 3m² + 15m + 12 [using (1)]
= 3 [k + m² + 5m + 4] = 3k’
where k’ = k + m² + 5m + 4 ∈ N as k, m ∈ N
which is multiple of 3.
⇒ P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 26.
n (n +1) (2n + 1) is divisible by 6.
Solution:
Let P (n) be the statement :
n (n + 1) (2n + 1) is divisible by 6.
Here P (1) means
1 . (1 + 1) (2 + 1) = 2 . 3 = 6 is divisible by 6, which is true,
Thus, P (1) is true.
Let us assume that P (n) is true for n = m
i.e. m (m + 1) (2m + 1) is divisible by 6.
i.e. m (m + 1) (2m + 1) = 6k, k ∈ N ………………..(1)
For P (m + 1) :
(m + 1) (m + 2) (2 (m + 1) + 1) = (m + 1) (m + 2) (2m + 3)
= m (m + 1) (2m + 1) + 2 (m + 1) (2m + 3) [using (1)]
= m (m + 1) (2m + 1) + 2m (m + 1) + 2 (m + 1) (2m + 3)
= 6k + 2 (m + 1) (m + 2m + 3) [using (1)]
= 6k + 6 (m + 1)²
= 6k’, which is divisible by 6.
where k’ = k + (m + 1)² ∈ N
Therefore, P (m + 1) is true.
Hence by M.I, P (n) is true for all n ∈ N.

Question 27.
x^{2n – 1} – 1 is divisible by (x — 1), x ≠ 1.
Solution:
Let P (n) be the statement :
x^{2n – 1} – 1 is divisible by (x — 1), x ≠ 1
Now P (1) means, x^{2 × 1} – 1
i.e. x – 1 is divisible by x – 1, x ≠ 1 which is true,
Thus, P (1) is true.
Let P (m) be true
i.e. x^{2m – 1} – 1 is divisible by x – 1.
x^{2m – 1} – 1 = (x – 1) f (x)
where f (x) be polynomial in x
x^{2m – 1} = (x – 1) f (x) + 1 …………………(1)
For P (m + 1) ;
x^{2 (m + 1) – 1} – 1 = 2^{2m + 1} – 1
= x^{2m – 1 + 2} – 1
= x^{2m – 1} x² – 1
= [(x – 1) f (x) + 1] x² – 1
= (x- 1) x² f (x) + x² – 1
= (x – 1) [x² f (x) + x + 1]
which is divisible by (x – 1)
⇒ P (m + 1) is true.
Thus, by principal of mathematucal induction P (n) is true for all n ∈ N.

Question 28.
3ⁿ > n for all n ∈ N.
Solution”
Let P (n) be the statement :
3ⁿ > n ∀ n ∈ N
Now P (1) means,
3¹ = 3 > 1
∴ P (1) is true.
Let P (m) be true i.e. 3^m > m ∀ m ∈ N ………………..(1)
For P (m + 1) :
3^{m + 1} = 3^m . 3 > 3.m ≥ m + 1 ∀ m ∈ N [using (1)]
[∵ m ≥ 1 > m + m ≥ m + 1
⇒ 3m ≥ 2m + 1 > m + 1 ∀ m ∈ N]
⇒ P (m + 1) is true
Hence by mathematical induction P(n) is true for all n ∈ N.

Question 29.
1 + 2 + 3 + ……………. + n < \(\frac{1}{8}\) (2n + 1)².
Solution:
Let P (n) be the statement :
1 + 2 + 3 + ……………. + n < \(\frac{1}{8}\) (2n + 1)²
Here P (1) means, 1 < \(\frac{9}{8}\) = \(\frac{1}{8}\) (2 . 1 + 1)²
∴ P (1) is true.
Let us assume that P (n) is true for n = m [using (1)]
∴ 1 + 2 + 3 + ………………. + m < \(\frac{1}{8}\) (2m + 1)²
Thus P (m + 1) ;
1 + 2 + 3 + ………………… + m + m + 1 < (2m + 1)² + m + 1 [using(1)]
= \(\frac{1}{8}\) [4m² + 4m + 1 + 8m + 8]
= \(\frac{1}{8}\) [4m² + 12m + 9]
= \(\frac{1}{8}\) (2m + 3)²
Therefore, P (m + 1) is true.
Hence by M.I, P (n) is true for all n ∈ N.

Question 30.
5 + 55 + 555 + ………………. to n terms = \(\frac{5}{81}\) (10^{n + 1} – 9n – 10).
Solution:
Let P (n) be the statement :
5 + 55 + 555 + …………………. to n terms = \(\frac{5}{81}\) (10^{n + 1} – 9n – 10)
Here P (1) :
5 = \(\frac{5}{81}\) (100 – 9 – 10)
= \(\frac{5}{81}\) × 81 = 5, which is true.
∴ P (1) is true.
Let us assume that P (n) is true for n = m.
i.e. 5 + 55 + 555 + …………………… m terms = \(\frac{5}{81}\) (10^{m + 1} – 9m – 10) ………………..(1)
Here (m + 1)th term = 555 ………………. (m + 1) terms
= 5 + 5 × 10 + 5 × 10² + ……………….. (m + 1) terms
= \(\frac{5\left(10^{m+1}-1\right)}{10-1}\)
= \(\frac{5}{9}\) (10^{m + 1} – 1) …………………….(2)
For P (m + 1) :
5 + 55 + 555 + …………………. (m + 1) terms = (5 + 55 + …………………. m terms) + (m + 1)th term
= \(\frac{5}{81}\) (10^{m + 1} – 9m – 10) + \(\frac{5}{9}\) (10^{m + 1} – 1) [using (1) and (2)]
= \(\frac{5}{81}\) [10^{m + 1} – 9m – 10 + 9 . 10^{m + 1} – 9]
= \(\frac{5}{81}\) [(1 + 9) 10^{m + 1} – 9 (m + 1) – 10]
= \(\frac{5}{81}\) [10^{m + 2} – 9 (m + 1) – 10]
∴ P (m + 1) is true.
Hence by M.I result is true for all n ∈ N.

Question 31.
If x is not an integral multiple of π, use induction to prove that sin x + sin 3x + sin 5x + …………………. + sin (2n – 1) x = \(\frac{\sin ^2 n x}{\sin x}\). for all n ∈ N.
Solution:
Let P (n) be the statement :
sin x + sin 3x + sin 5x + …………………. + sin (2n – 1) x = \(\frac{\sin ^2 n x}{\sin x}\) ∀ n ∈ N
Now P (1) means,
sin x = \(\frac{\sin ^2 x}{\sin x}\) = sin x, which is true
⇒ P(1) is true.
Let P (m) is true
i.e. sin x + sin 3x + sin 5x + …………………. + sin (2m- 1) x = \(\frac{\sin ^2 m x}{\sin x}\) ……………….(1)
For P (m + 1) ;
[sin x + sin 3x + sin 5x + ………………………. + sin (2m – 1) x] + sin (2m + 1) x

⇒ P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.