Question

Question 15 Prove the following by using the principle of mathematical induction for all n∈N: 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = [n(2n-1)(2n+1)] / 3

Class X1 - Maths -Principle of Mathematical Induction Page 95

Answer 1

1² + 3² + 5² + ......+ (2n-1)² = n(2n-1)(2n+1)/3
Let P(n): 1² + 3² + 5² + . .... ..+(2n-1)² = n(2n-1)(2n+1)/3

step1:- for n = 1
P(1): 1² = 1(2-1)(2+1)/3 = 3/3 = 1
it's true.

step2:- for n = K
P(K): 1² + 3² + 5² +.....+(2k-1)² = K(2K-1)(2K-1)/3 _________(1)

step3:- for n= k+1
P(K+1): 1² + 3² + 5² + .......+(2k+1)² = (k+1)(2K+1)(2k+3)/3

from eqn (1)
1² + 3² + 5² + .. ......+ (2K -1)² = K(2K-1)(2K+1)/3
add ( 2K+1)² both sides,
1² + 3² + 5² + ......+ (2k-1)² + (2K+1)² = K(2k-1)(2k+1)/3 + (2K+1)²
= (2k+1)/3{2k² - K + 6K + 3}
= (2K+1)(2K² + 5K + 3)/3
= (2K+1)(2K+3)(K+1)/3
= [(k+1){2(k+1)-1}{2(K+1)+1}]/3

hence, p(k+1) is true when P(K) is true , from the principle of mathematical induction , statement is true for all real numbers.

Answer 2

Answer:

Step-by-step explanation:

Let P(n) be the given statement, i.e., P(n) : (2n + 1) < 2n

for all natural

numbers, n ≥ 3. We observe that P(3) is true, since

2.3 + 1 = 7 < 8 = 23

Assume that P(n) is true for some natural number k, i.e., 2k + 1 < 2k

To prove P(k + 1) is true, we have to show that 2(k + 1) + 1 < 2k+1. Now, we have

2(k + 1) + 1 = 2 k + 3

= 2k + 1 + 2 < 2k + 2 < 2k

. 2 = 2k + 1

.

Thus P(k + 1) is true, whenever P(k) is true.

Hence, by the Principle of Mathematical Induction P(n) is true for all natural

numbers, n ≥ 3