Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 4 Principle of Mathematical Induction Ex 4.1 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction Ex 4.1
Question 1.
Prove the following by using the principle of mathematical induction for all n ∈ N:
1 + 3 + 32 + …………. + 3n – 1 = \(\frac{3^{n}-1}{2}\)
Answer.
Step I: Let P( n) be the given statement.
i.e., 1 + 3 + 32 + …………. + 3n – 1 = \(\frac{3^{n}-1}{2}\)
Step II: For n = 1, we have
LHS = 31 – 1
= 30 = 1
Thus, P(n)is true form = 1
Step III: For n = k, assume that P( k) is true. Then,
1 + 3 + 32 + …………. + 3k – 1 = \(\frac{3^{k}-1}{2}\)
Step IV: For n = k + 1,
We have to show that P(k + 1) is true, whenever P(k) is true.
i.e., 1 + 3 + 32 + …………… + 3k – 1 + 3k = \(\frac{3^{k+1}-1}{2}\)
L.H.S. = 1 + 3 + 32 + …………… + 3k – 1 + 3k
= \(\frac{3^{k+1}-1}{2}\) + 3k
= \(\frac{3^{k}-1+2.3^{k}}{2}=\frac{3^{k}(1+2)-1}{2}\)
= \(\frac{3^{k} \cdot 3-1}{2}=\frac{3^{k+1}-1}{2}\)
= R.H.S
Thus, P(k + 1) is true, whenever P(k) is true.
Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.
Question 2.
Prove the following by using the principle of mathematical induction for all n ∈ N:
13 + 23 + 33 + …………. + n3 = \(\left(\frac{n(n+1)}{2}\right)^{2}\).
Answer.
Let the given statement be P(n), i.e.,
P(n) = 13 + 23 + 33 + …………. + n3 = \(\left(\frac{n(n+1)}{2}\right)^{2}\).
For n = 1, we have
P(1) : 13 = 1
= \(\left(\frac{1(1+1)}{2}\right)^{2}=\left(\frac{12}{2}\right)^{2}\)
= 12 = 1 which is true.
Let P(k) be true for some positive integer k, i.e.,
13 + 23 + 33 + …………. + k3 = \(\left(\frac{k(k+1)}{2}\right)^{2}\)
We shall now prove that P(k+1) is true.
Consider 13 + 23 + 33 + …………. + k3 = \(\left(\frac{k(k+1)}{2}\right)^{2}\)
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 3.
Prove the following by using the principle of mathematical induction for all n ∈ N:
\(1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\ldots+\frac{1}{(1+2+3+\ldots n)}=\frac{2 n}{(n+1)}\).
Answer.
Step I : Let the gven statement be P(n).
Step II: For n = k, assume the P(k) is true.
i.e., \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{2}{k(k+1)}=\frac{2 k}{k+1}\) ……………(i)
Step IV: For n = k + 1, we have to show that P(k + 1) is true whenever, P(k) is true.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, the statement is true for all natural numbers n.
Question 4.
Prove the following by using the principle of mathematical induction for ail n ∈ N:
1 . 2 . 3 + 2 . 3 . 4 + …………. + n (n + 1) (n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
Answer.
Let the given statement be P(n),
i.e., P(n) :
1 . 2 . 3 + 2 . 3 . 4 + …………. + n (n + 1) (n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
For n = 1, we have
P(1): 1 . 2 . 3 = 6
= \(\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1.2 .3 .4}{4}\) = 6, which is true.
Let P(k) be true for some positive integer k, i.e.,
1 . 2 . 3 + 2 . 3 . 4 + …………. + k (k + 1) (k + 2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\) …………….(i)
We shall now prove that P(k + 1) is true.
Consider 1 . 2 . 3 + 2 . 3 . 4 + …………. + k (k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)
= {1 . 2 . 3 + 2 . 3 . 4 + …………… + k (k + 1) (k + 2)} + (k + 1) (k + 2) (k + 3)
= \(\frac{k(k+1)(k+2)(k+3)}{4}\) + (k + 1) (k + 2) (k + 3) [using eq. (i)]
= (k + 1) (k + 2) (k + 3) [\(\frac{k}{4}\) + 1]
= \(\frac{(k+1)(k+2)(k+3)(k+4)}{4}\)
= \(\frac{(k+1)(k+1+1)(k+1+2)(k+1+3)}{4}\)
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 5.
Prove the following by using the principle of mathematical induction for all n ∈ N:
1 . 3 + 2 . 32 + 3 . 33 + ……….. + n . 3n = \(\frac{(2 n-1) 3^{n+1}+3}{4}\)
Answer.
Let the given statement be P(n), i.e.,
P(n) : 1 . 3 + 2 . 32 + 3 . 33 + …………… + n . 3n = \(\frac{(2 n-1) 3^{n+1}+3}{4}\)
For n = 1,
we have P(1) : 1 . 3 = 3
= \(\frac{(2.1-1) 3^{1+1}+3}{4}=\frac{3^{2}+3}{4}=\frac{12}{4}\) = 3, which is true.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 6.
Prove the following by using the principle of mathematical induction for all n ∈ N:
1 . 2 + 2 . 3 + 3 . 4 + ……………. + n . (n + 1) = \(\left(\frac{n(n+1)(n+2)}{3}\right)\)
Answer.
Let the given statement be P(n), i.e.,
P(n) : 1 . 2 + 2 . 3 + 3 . 4 + ……………. + n . (n + 1) = \(\left(\frac{n(n+1)(n+2)}{3}\right)\)
For n = 1, we have
P(1) : 1 . 2 = 2
= \(\frac{1(1+1)(1+2)}{3}=\frac{1 \cdot 2 \cdot 3}{3}\) = 2, which is true.
Let P(k) be true for some positive integer k, i.e.,
1 . 2 + 2 . 3 + 3 . 4 + ………………. + k . (k + 1) = \(\left(\frac{k(k+1)(k+2)}{3}\right)\) …………….(i)
We shall now prove that P(k + 1) is true.
Consider 1 . 2 + 2 . 3 + 3 . 4 + …………. + k . (k + 1) + (k + 1) . (k + 2)
= [1 . 2 + 2 . 3 + 3 . 4 + …………….. + k . (k + 1)] + (k + 1) . (k + 2)
= \(\frac{k(k+1)(k+2)}{3}\) + (k + 1) (k + 2)
= (k + 1) (k + 2) + (\(\frac{k}{3}\) + 1)
= \(\frac{(k+1)(k+2)(k+3)}{3}\)
= \(\frac{(k+1)(k+1+1)(k+1+2)}{3}\)
Thus, P(k+1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 7.
Prove the following by using the principle of mathematical induction for all n ∈ N:
1 . 3 + 3 . 5 + 5 . 7 + … + (2n – 1) (2n + 1) = \(\frac{n\left(4 n^{2}+6 n-1\right)}{3}\)
Answer.
Let the given statement be P(n), i.e.,
P(n) : 1 . 3 + 3 . 5 + 5 . 7 + … + (2n – 1) (2n + 1) = \(\frac{n\left(4 n^{2}+6 n-1\right)}{3}\)
For n = 1, we have
P(1) : 1 . 3 = 3
= \(\frac{1\left(4.1^{2}+6.1-1\right)}{3}=\frac{4+6-1}{3}=\frac{9}{3}\) = 3, which is true.
Let P(k) be true for some positive integer k, Le.,
1 . 3 + 3 . 5 + 5 . 7 + ………………. + (2k – 1) (2k + 1) = \(\frac{k\left(4 k^{2}+6 k-1\right)}{3}\) ……………(i)
We shall now prove that P(k + 1) is true.
Consider (1 . 3 + 3 . 5 + 5 . 7 + … + (2k – 1) (2k + 1) +{2 (k + 1) – 1} {2(k + 1) + 1}
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, .statement P(n) is true for all natural numbers i.e., n.
Question 8.
Prove the following by using the principle of mathematical induction for all n eN:
1.2 + 2.22 + 3.22 + …………. + n.2n = (n – 1) 2n + 1 + 2
Answer.
Let the given statement be P(n), i.e.,
P(n) : 1.2 + 2.22 + 3.22 + …………. + n.2n = (n – 1) 2n + 1 + 2
For n = 1, we have
P(1) : 1 . 2 = 2
= (1 – 1) 2k + 1 + 2
= 0 + 2 = 2, which is true.
Let P(k) be true for some positive integer k, i.e.,
1.2 + 2.22 + 3.22 + …………. + k.2k = (k – 1) 2k + 1 + 2 ……………(i)
We shall now prove that P(k + 1) is true.
Consider {1.2 + 2.22 + 3.22 + …………. + k.2k} + (k – 1) 2k + 1 + 2
= (k – 1)2k + 1 + 2 + (k + 1)2k + 1
= 2k + 1 {(k – 1) + (k + 1)} + 2
= 2k + 1 . 2k + 2
= k . 2(k + 1) + 1 + 2
= {(k + 1) – 1} 2(k + 1) + 1 + 2
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 9.
Prove the following by using the principle of mathematical induction for all n ∈ N:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}\)
Answer.
Let the given statement be P(n), i.e.,
P(n) : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}\)
For n = 1, we have
P(1) : \(\frac{1}{2}\)
= 1 – \(\frac{1}{2^{1}}\)
= \(\frac{1}{2}\), which is true.
Let P(k) be true for some positive integer k, i.e.,
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{k}}=1-\frac{1}{2^{k}}\)
We shall now prove that P(k + 1) is true.
Consider \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{k}}\right)+\frac{1}{2^{k+1}}\)
= \(\left(1-\frac{1}{2^{k}}\right)+\frac{1}{2^{k+1}}\)
= \(1-\frac{1}{2^{k}}+\frac{1}{22^{k}}=1-\frac{1}{2^{k}}\left(1-\frac{1}{2}\right)\)
= \(1-\frac{1}{2^{k}}\left(\frac{1}{2}\right)=1-\frac{1}{2^{k+1}}\)
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 10.
Prove the following by using the principle of mathematical induction for all n ∈ N:
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{(6 n+4)}\)
Answer.
Let the given statement e P(n), i.e.,
P(n) : \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{(6 n+4)}\)
For n = 1, we have P(1) = \(\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6.1+4}=\frac{1}{10}\), which is true.
Let P(k) be true for some positive integer k, i.e.,
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\ldots+\frac{1}{(3 k-1)(3 k+2)}=\frac{k}{6 k+4}\) …………..(i)
We shall now prove that P(k + 1) is true.
Consider
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 11.
Prove the following by using the principle of mathematical induction for all n ∈ N:
\(\frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}\)
Answer.
Let the given statement p(n), i.e.,
P(n) : \(\frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}\)
For n = 1, we have
P(1) : \(\frac{1}{1.2 .3}=\frac{1 .(1+3)}{4(1+1)(1+2)}=\frac{1.4}{42.3}=\frac{1}{1.2 .3}\), which is true.
Let P(k) e true for some positive integer k, i.e.,
\(\frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)}\) ………………(i)
we shall now prove that P(k + 1) is true.
Consider
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 12.
Prove the following by using the principle of mathematical induction for all n ∈ N:
a + ar + ar2 + …………… + arn – 1 = \(\frac{a\left(r^{n}-1\right)}{r-1}\)
Answer.
Let the given statement be P(n), i.e.,
P(n) : a + ar + ar2 + …………… + arn – 1 = \(\frac{a\left(r^{n}-1\right)}{r-1}\)
For n = 1, we have
P(1) : a = \(\frac{a\left(r^{1}-1\right)}{(r-1)}\) = a, which is true.
Let P(k) be true for some positive integer k, i.e.,
a + ar + ar2 + …………… + ark – 1 = \(\frac{a\left(r^{k}-1\right)}{r-1}\) …………..(i)
We shall now prove that P(k + 1) is true.
Consider
a + ar + ar2 + …………… + ark – 1} + ar(k + 1) – 1 = \(\frac{a\left(r^{k}-1\right)}{r-1}\) + ark [Using eQuestion (i)]
= \(\frac{a\left(r^{k}-1\right)+a r^{k}(k-1)}{r-1}\)
= \(\frac{a\left(r^{k}-1\right)+a r^{k+1}-a r^{k}}{r-1}\)
= \(\frac{a r^{k}-a+a r^{k+1}-a r^{k}}{r-1}=\frac{a r^{k+1}-a}{r-1}\)
= \(\frac{a\left(r^{k+1}-1\right)}{r-1}\)
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 13.
Prove the following by using the principle of mathematical induction for all n ∈ N:
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \cdots\left(1+\frac{(2 n+1)}{n^{2}}\right)\) = (n + 1)2
Answer.
= (k + 1)2 + 2k + 3
= k2 + 2k + 1 + 2k + 3
= k2 + 4k + 4
= (k + 2)2
= (k + 1 + 1)2
= R.H.S.
So, P( k + 1) is true, whenever P( k) is true.
Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.
Question 14.
Prove the following by using the principle of mathematical induction for all n ∈ N:
\(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{n}\right)\) = (n + 1)
Answer.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 15.
Prove the following by using the principle of mathematical indcution for all n ∈ N:
12 + 32 + 52 + … + (2n – 1)2 = \(\frac{n(2 n-1)(2 n+1)}{3}\)
Answer.
Let the given statement be P(n), i.e.,
P(n) = 12 + 32 + 52 + … + (2n – 1)2 = \(\frac{n(2 n-1)(2 n+1)}{3}\)
For n = 1, we have
P(1) = 12 = 1
= \(\frac{1(2.1-1)(2.1+1)}{3}=\frac{1.1 .3}{3}\) = 1, which is true.
Let P(k) be true for some positive integer k, i.e.,
P(k) = 12 + 32 + 52 + … + (2k – 1)2 = \(\frac{k(2 k-1)(2 k+1)}{3}\)
We shall now prove that P(k + 1) is true.
Consider {12 + 32 + 52 + ………… + (2k – 1)2 } + {2(k + 1) – 1}2
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical indcution, statement P{n) is true for all natural numbers i.e., n.
Question 16.
Prove the following by using the principle of mathematical induction for all n ∈ N:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{(3 n+1)}\)
Answer.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 17.
Prove the following by using the principle of mathematical induction for all n ∈ N:
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}\)
Answer.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 18.
Prove the following by using the principle of mathematical induction for all n ∈ N:
1 + 2 + 3 + ………… + n < \(\frac{1}{8}\) (2n + 1)2
Ans.
Let the given statement be P(n), i.e.,
P(n) : 1 + 2 + 3 + ………… + n < \(\frac{1}{8}\) (2n + 1)2
It can be noted that P(n) is true for n = 1 since 1 < \(\frac{1}{8}\) (2.1 + 1)2 = \(\frac{9}{8}\)
Let P(k) be true for some positive integer k, i.e.,
1 + 2 + 3 + ………… + k < \(\frac{1}{8}\) (2k + 1)2
We shall now prove that P(k+1) is true whenever P(k) is true.
Consider
Thus P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 19.
Prove the following by using the principle of mathematical induction for all n Answer.
Let the given statement be P(n), i.e.,
P(n) : n (n + 1) (n + 5) which is a multiple of 3.
It can be noted that P(n) is true for n=l since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3.
Let P(k) be true for some positive integer k, i.e., k (k + 1) (k + 5) is a multiple of 3.
∴ k (k + 1) (k + 5) = 3m, where m ∈ N ……………….(i)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider (k + 1) {(k +1 ) + 1} {(k + 1) + 5}
= (k + 1) (k + 2) {(k + 5) + 1}
= (k + 1) (k + 2) (k + 5) + (k + 1) (k + 2)
= {k (k + 1) (k + 5) + 2 (k + 1) (k + 5)} + (k +1) (k + 2)
= 3m + (k + 1) {2 (k + 5) + (k + 2)}
= 3m + (k + 1) {2k +10 + k + 2}
= 3m + (k + 1) (3k + 12)
= 3m + 3 (k + 1) (k + 4)
= 3 {m + (k + 1) (k + 4)}
= 3 × q, where, q = {m + (k +1) (k + 4)} is some natural number.
Therefore, (k +1) {(k +1) +1} {(k +1) + 5} is a multiple of 3.
Thus, P(k + 1) is trtie whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 20.
Prove the following by using the principle of mathematical induction for all n GN : 102″ 1 + 1 is divisible by 11.
Answer.
Let the given statement be P(n), i.e.,
P(n) : 102n – 1 + 1 is divisible by 11.
It can be observed that P(n) is true for n = 1 since
P(1) = 102 . 1 – 1 + 1 = 11, which is divisible by 11.
Let P(k) be true for some positive integer k, i.e.,
102 – 1 + 1 is divisible by 11.
∴ 102k – 1 + 1 = 11 m, where m ∈ N …………..(i)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider 102k + 1 + 1
= 102k + 2 – 1 + 1
= 102k + 1 + 1
= 102 (102k – 1 + 1 – 1) + 1
= 102 (102k – 1 + 1) – 102 + 1
= 102 . 11m – 100 + 1 [using eq. (i)]
= 100 × 11m – 99
= 11 (100m – 9)
= 11r, where r = (100m – 9) is some natural number.
Therefore, 102(k + 1) – 1 + 1 is divisible by 11.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 21.
Prove the following by using the principle of mathematical induction for all n ∈ N: x2n – y2n is divisible by x + y.
Answer.
Let the given statement be P(n), i.e.,
P(n) : x2n – y2n is divisible by x + y.
It can be observed that P(n) is true for n = 1.
This is so because x2 × 1 – y2 × 1
= x2 – y2
= (x + y) (x – y) is divisible by (x + y).
Let P(k) be true for some positive integer k, i.e., x2k – y2k is divisible by x + y.
∴ x2k – y2k = m (x + y), where m ∈ N ………….(i)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider x2(k + 1) – y2 (k + 1)
= x2k . x2 – y2k . y2
= x2 (x2k – y2k + y2k) – y2k. y2
= x2 {m (x + y) + y2k} – y2k . y2 [using eq. (i)]
= m (x + y) x2 + y2k x2 – y2 . y2
= m (x + y) x2 + y2k (x2 – y2)
= m (x + y) x2 + y2k (x + y)(x – y)
= (x + y) {mx2 + y2k (x – y)}, which is a factor of (x + y).
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e.,n.
Question 22.
Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n – 9 is divisible by 8.
Answer.
Step I: Let P(n) be the given statement. Then,
P(n): 32n + 2 – 8n – 9 is divisible by 8.
Step II: For n = 1, we have
P(1) : 32 + 2 – 8 . 1 – 9
= 34 – 8 – 9
= 81 – 17
= 64, which is divisible by 8.
Step III: For n = k, assume that P(k) is true.
Then, P(k): 32k + 2 – 8k – 9 is divisible by 8.
⇒ 32k + 2 – 8k – 9 where m ∈ N.
⇒ 32k + 2 = 8m + 8k + 9 …………..(i)
Step IV: For n = k + 1, we have to show that P(k +1) i.e.,
32(k + 1) + 2 + 2 – 8(k + 1) – 9 is divisible by 8.
Now, P(k + 1): 32(k + 1) + 2 – 8(k + 1) – 9
= 32k + 2 + 2 – 8k – 8 – 9
= 32 – 32k + 2 – 8k – 17
= 9 (8m + 8k + 9) – 8k – 17 [from eq. (i)]
= 72m + 72k + 81 – 8k – 17
= 72m + 64 k + 64 = 8 [9m + 8k + 8] which is divisible by 8.
So, P(k + 1) is true whenever PCk) is true.
Hence, by principle of mathematical induction, P(n) is true for all n ∈ N.
Question 23.
Prove the following by using the principle of mathematical indcution for all it e N: 41″ – 14″ is a multiple of 27.
Answer.
Let the given statement be P(n), i.e.,
P(n) : 41n – 14n is a multiple of 27.
It can be observed that P(n) is true for n=l since 411 – 141 = 27
Let P(k) be true for some positive integer k, i.e.,
41k – 14k is a multiple of 27
∴ 41k – 14k = 27m, where m ∈ N …………….(i)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider 41k + 1 – 14k + 1
= 41k . 41 – 14k . 14
= 41 (41k – 14k + 14k) – 14k .14
= 41 (41k – 14k) + 41 . 14k – 14k . 14
= 41 . 27m + 14k (41 – 14)
= 41 . 27m + 27 . 14k = 27 (41m – 14k)
= 27 × r, where r = (41m – 14k) is a natural number.
Therefore, 41k + 1 – 14k + 1 is a multiple of 27.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
Question 24.
Prove the following by using the principle of mathematical induction for all n ∈ N:
(2n + 7) < (n + 3)2
Answer.
Let the given statement be P(n), i.e., P(n) : (2n + 7) < (n + 3)2
It can be observed that P(n) is true for n = 1 since
2 . 1 + 7 = 9 < (1 + 3)2 = 16, which is true.
Let P(k) be true for some positive integer k, i.e.,
(2k + 7) < (k + 3)2 …………..(i)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider {2 (k +1) + 7} = (2k + 7) + 2
{2 (k + 1) + 7} = (2k + 7) + 2 < (k + 3)2 + 2 [using eq. (i)]
2 (k + 1) + 7 < k2 + 6k + 9 + 2
2 (k + 1) + 7 < k2 + 6k + 11
Now, k2 + 6k + 11 < k2 + 8k + 16
2 (k + 1) + 7 < (k + 4)2
(k + 1) + 7 < {(k + 1) + 3}2
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
