Question

n(4n² +6n-D
7. 1.3 +3.5 +5.7 +...+ (2n-1) (2n+1) = "
(
1) On+II​

Answer 1

Explanation:

1.3 + 3.5 + 5.7 + .......+(2n-1)(2n+1) = n(4n²+6n-1)/3

Let p(n): 1.3 + 3.5 + 5.7 + ......+(2n-1)(2n+1) = n(4n²+6n-1)/3

step1:- for n = 1

P(1): 1.3 = 1.(4.1²+6.1-1)/3

= 1.(4+6-1)/3 = 1.3

It's true.

step2:- for n = k

P(k): 1.3 + 3.5 + 5.7 +......+(2k-1)(2k+1) = k(4k²+6k-1)/3 -------(1)

step3:- for n = k+1

P(k+1) :1.3 + 3.5 + 5.7 +....+(2k-1)(2k+1) + (2k+1)(2k+3) = (k+1){4(k+1)²+6(k+1)-1}/3

from eqn (1)

1.3 + 3.5 + 5.7 + .....+(2k-1)(2k+1)+ = k(4k²+6k-1)/3

add (2k+1)(2k+3) both sides ,

1.3 + 3.5 + 5.7 + .........+(2k-1)(2k+1)+(2k+1)(2k+3)= k(4k² + 6k-1)/3 + (2k+1)(2k+3)

= {k(4k²+6k-1)+3(4k²+8k+5)}/3

= (4k³ + 6k²- k + 12k²+24k+15)/3

= (4k³ + 18k² + 23k + 9)/3

= (4k³ + 4k² +14k²+14k+9k+9)/3

= (k+1)(4k²+14k+9)/3

= (k+1) {4(k+1)² +6(k+1)-1}/3

Hence, p(k+1) is true when p(k) is true . from the mathematical induction, statement is true for all natural numbers.