Question

Prove that,

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

Answer 1

When n = 1, we have:
L.H.S. = (2-1)*(2+1) = 1*3 = 3
R.H.S. = 1(4+6-1)/3 = 9/3 = 3

L.H.S. = R.H.S., so P(n) holds true for n = 1.

Assume that P(n) holds true for n = k, where k is some positive integer.
Now, we need to show that P(n) holds true for n = k + 1 as well.

When n = k + 1, we have:
L.H.S.
= 1*3 + 3*5 + 5*7 + ... + (2k-1)*(2k+1) + (2k+1)*(2k+3)
= [1*3 + 3*5 + 5*7 + ... + (2k-1)*(2k+1)] + (2k+1)*(2k+3)
= k*(4k^2+6k-1)/3 + (2k+1)*(2k+3) [By induction hypothesis]
= k*(4k^2+6k-1)/3 + 3*(2k+1)*(2k+3)/3
= [k(4k^2+6k-1) + 3(2k+1)(2k+3)]/3
= [4k^3 + 6k^2 - k + 12k^2 + 24k + 9]/3
= [4k^3 + 18k^2 + 23k + 9]/3
= (k+1)[4(k+1)^2 + 6(k+1) - 1]/3
= R.H.S.

Thus, P(n) holds true for n = k + 1.
Hence, by mathematical induction, P(n) holds true for all positive integer n.