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## Karnataka 1st PUC Maths Question Bank Chapter 4 Principle of Mathematical Induction

Question 1.

Using the principle of mathematical induction, prove that

\(\text { (1) } 1+2+3+\cdots+n=\frac{n(n+1)}{2}\), for all n∈N

Answer:

which is P(k +1)

Thus, P(k)⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 2.

1 + 3 + 5 + 7 + …….. + (2n – 1) = n^{2}, for all n∈N

Answer:

Let P(n):1 + 3 + 5 + 7+……….. +(2n-1) = n^{2
}For n = 1, LHS = 1, RHS = 1^{2} = 1

∴ LHS = RHS

∴ F(1) is true.

Let upwards assume P(k) is true for some k∈N

i.e., 1 + 3 + 5 + …. + (2K- 1) = k^{z
}Adding (k + 1)^{th} term = 2K + 1, on both sides, we get,

1 + 3 + 5 +……. +(2k -1)(2k + l) = k^{2} + 2k + 1

= (k +1)^{2
}which is P(k +1)

Thus, P(k)⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 3.

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

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 4.

\(1^{3}+2^{3}+3^{3}+\ldots \ldots+n^{3}=\left[\frac{n(n+1)}{2}\right]^{2} \) for all n∈N

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 5.

\(1 \cdot 2+2 \cdot 3+\dots+n(n+1)=\frac{n(n+1)(n+2)}{3}\) for all n∈N

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 6.

\( \begin{aligned} &1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+\dots+n(n+1)(n+2)&=\frac{n(n+1)(n+2)(n+3)}{4},

\text { for all } n \in N \end{aligned}\)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 7.

\( \cfrac{1}{1 \cdot 2}+\cfrac{1}{2 \cdot 3}+\cfrac{1}{3 \cdot 4}+\dots+\cfrac{1}{n(n+1)}=\cfrac{n}{n+1}\) for all n∈N

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 8.

\(\begin{aligned} &\cfrac{1}{1 \cdot 2 \cdot 3}+\cfrac{1}{2 \cdot 3 \cdot 4}+\cfrac{1}{3 \cdot 4 \cdot 5}+\dots+\cfrac{1}{n(n+1)(n+2)}&=\cfrac{n(n+3)}{4(n+1)(n+2)}, \text { for all } n \in N \end{aligned}\)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 9.

\( \begin{aligned}&\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\cdots \cdot+\\ &\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{6 n+4} \end{aligned}\)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 10.

\(\begin{aligned} &\cfrac{1}{1 \cdot 4}+\cfrac{1}{4 \cdot 7}+\cfrac{1}{7 \cdot 10}+\cdots+\\ &\cfrac{1}{(3 n-2)(3 n+1)}=\cfrac{n}{3 n+1} \end{aligned} \)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 11.

\(\begin{aligned} &\cfrac{1}{3 \cdot 5}+\cfrac{1}{5 \cdot 7}+\cfrac{1}{7 \cdot 9}+\cdots+\\ &\cfrac{1}{(2 n+1)(2 n+3)}=\cfrac{n}{3(2 n+3)} \end{aligned}\)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 12.

\(\begin{aligned} &1+\cfrac{1}{1+2}+\cfrac{1}{1+2+3}+\dots+\\ &\cfrac{1}{1+2+3+\dots+n}=\cfrac{2 n}{n+1} \end{aligned} \)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 13.

\(\begin{aligned}&1 \cdot 3+3 \cdot 5+5 \cdot 7+\dots+(2 n-1)(2 n+1)&=\cfrac{n\left(4 n^{2}+6 n-1\right)}{3}, \text { for all } n \in N \end{aligned}\)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 14.

\(1+3+3^{2}+\cdots+3^{n-1}=\cfrac{3^{n}-1}{2} \)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 15.

\(\cfrac{1}{2}+\cfrac{1}{4}+\cfrac{1}{8}+\dots+\cfrac{1}{2^{n}}=1-\cfrac{1}{2^{n}} \)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 16.

\( a+a r+a r^{2}+\cdots+a r^{n-1}=\cfrac{a\left(r^{n}-1\right)}{r-1}\)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 17.

\( 1 \cdot 2+2 \cdot 2^{2}+3 \cdot 2^{3}+\cdots+n 2^{n}=(n-1) 2^{n+1}+2\)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 18.

\( \begin{aligned} &1 \cdot 3+2 \cdot 3^{2}+3 \cdot 3^{2}+\cdots+n \cdot 3^{n}&=\cfrac{(2 n-1) 3^{n+1}+3}{4}, \text { for all } n \in N \end{aligned} \)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 19.

\(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{n}\right)=n+1\) for all n∈N

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 20.

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

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 21.

\(1^{2}+3^{2}+5^{2}+\cdots \cdot \cdot+(2 n-1)^{2}=\cfrac{n(2 n-1)(2 n+1)}{3}\)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 22.

Prove that 2^{n}> n for all positives integers

Answer:

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 23.

\( 1+2+3 \cdots+n<\frac{1}{8}(2 n+1)^{2}\)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 24.

\(1^{2}+2^{2}+3^{2}+\cdots+n^{2}>\frac{n^{3}}{3}, n \in N \)

Answer:

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 25.

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

Let P(n): (2n + 7) < (n + 3)^{2
}For = 1, (2 + 7)<(1+ 3)^{2} ⇒9<16 is true.

∴ P(1) is true.

Let us assume P(k) is true for some k∈ N

i.e., (2k+ 1) < (k + 3)^{2}

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

= (2k + 7) + 2

< (k: + 3)^{2} + 2 = k^{2} + 6k + 9 + 2

<(k + 4)^{2
}∴ P(k +1) is true.

Hence, by mathematical induction, P(n) is true for all n∈N

Question 26.

7^{2n} – 3^{2n} is divisible by 4.

Answer:

Let P(n): 7^{n} – 3^{n} is divisible by 4.

For n = 1, P(1): 7^{1} – 3^{1} = 4 which is divisible by 4.

∴ P(1) is true.

Let us assume P(k) is true for some k∈N

i.e., 7^{2n} – 3^{2n }is divisible by 4.

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 27.

x^{2n} – y^{2n} is divisible by x+y.

Answer:

Let P(n): x^{2n} – y^{2n} is divisible by x + y

For n = 1, P(1): x^{2} – y^{2} is divisible by x + y, which is true.

∴ P(1) is true.

Let us assume P(k) is true for some k∈ N

i.e., x^{2k} – y^{2k} is divisible by x + y

Let x^{2k} – y^{2k} = (x + y)d, where d∈N

which is P(k +1)

Thus, P(k) ⇒ P(k +1).

Hence, by mathematical induction P(n) is true for all n∈N

Question 28.

10^{2n}^{-1} +1 is divisible by

Answer:

Let P(n): 10^{2n-}^{1} +1, is divisible by 11.

For n = 1, P(1):10^{2-1} +1, is divisible by 11, which is true.

P(1) is true.

Let us assume P(k) is true for some k∈N.

i.e., 10^{2n}^{-1}+1 is divisible by 11.

⇒10^{2}^{-1} +1 = 1 W, d∈N.

Consider 10^{2(k+1)-}^{1}+1 = 10^{2k}^{+1} +1

= 10^{2k}^{-1 }10^{2}+1 = (11d-1)10^{2}+1

= 11d(10^{2})-100 + 1 = 11(100d)-99

= 11(100d-9)

∴ 10^{2(k+1)-1} +1 is divisible by 11.

Thus, P(k) ⇒ P(k +1)

Hence, by mathematical induction, P(n) is true for all n∈N

Question 29.

3^{2n}^{+2} -8n-9 is divisible by 8.

Answer:

Let P(n): 3^{2n}^{+2} -8n-9 is divisible by 8.

For n = 1, P(1): 3^{2+2} – 8(2) – 9 = 64 is divisible by 8, which is true.

P( 1) is true.

Let us assume P(k) is true for some k∈N

i.e., 3^{2k+}^{2} -8k -9 is divisible by 8.

Let 3^{2k+2}-8k-9 = 8d, d∈N ………………(1)

Consider 3^{2(k}^{+1)+2} – 8(k +1) – 9 = 3^{2k}^{+4} – 8k – 8 – 9

= 3^{2k+2} -3^{2}-8(k-1) = (8d + 8k + 9)9-8k-17 using (1)

= (8d + 8k)9 + 81-8k-17

= 8(d + k)9 + 64-8k = 8[9(d + k) + 8-k]

which is divisible by 8.

Thus, P(k) ⇒ P(k +1)

Hence, by mathematical induction, P(n) is true for all n∈N

Question 30.

2.7^{n} + 3.5^{n} – 5 is divisible by 24 for all n∈N

Answer:

Let P(n): 2.7^{n} + 3.5^{n} – 5 is divisible by 24.

For n = 1; P(1): 2.7^{n} + 3.5^{n} – 5 = 24 is divisible by 24. ‘

∴ P(1) is true.

Let us assume, P(k) is true for some k∈N

i.e P(n): 2.7^{k} + 3.5^{k} – 5 is divisible by 24.

Let 2.7^{k} + 3.5^{k} – 5=24d,where d∈N

Thus, P(k)⇒P(k+ l)

Hence, by mathematical induction, P(n) is true for all n∈N

Question 31.

41^{n} -14^{n} is a multiple of 27.

Answer:

Let P(n): 41^{n} -14^{n} is a multiple of 27.

For n = 1, P(1): 41′-14′ = 47 is a multiple of 27.

∴ P(1) is true.

Let us assume P(k) is true for some k∈N

Thus, P(k)⇒P(k+ 1)

Hence, by mathematical induction, P(n) is true for all n∈N

Question 32.

n(n +1)(n + 5) is a multiple of 3.

Answer:

Let P(n): n(n + 1)(n + 5) is a multiple of 3.

For n = 1, P(1): 1(1 +1)(1 + 5) = 12 is a multiple of 3.

∴ P(1) is true.

Let us assume P(k) is true for some n∈N

i.e., P(k):k(k + 1)(k + 5) is a multiple of 3.

Let k(k + 1)(k + 5) = 3d

Consider (k + 1)(k + 2)(k + 6)

= (k + 2)(k + 1)(k + 5 + 1)

= k(k + 1)(k + 5 + 1) + 2(k + 1)(k + 5 + 1)

= k(k +1 )(k + 5) + k(k +1) + 2(k +1 )(k + 6)

Question 33.

\((a b)^{n}=a^{n} b^{n}\)

Answer:

Question 34.

(1 + x)^{n} ≥ (1 + nx)for all natural number n, where x>-l.

Answer:

Let P(n): (1 + x)^{n} ≥ (1 + nx), for x >-1

For n = 1; P(1): (1 + x)^{1} ≥ (1 + x) which is true.

Let us assume P(k) is true for some k∈N

i.e., (1 + x)^{k} ≥ (1 + kx)

Consider (1 + x)^{k+1} = (1 + x)^{k} (1 + x)

≥ (1 + kx)(1 + x)

= (l + kx + x + kx^{2})

≥(1 + (k + 1)x)

∵ kx^{2}≥0.

which is P(k + 1)

Thus, P(k) ⇒ P(k +1)

Hence, by mathematical induction, P(n) is true for all n∈N