Find the sum of:
1⋅3⋅5⋅7+3⋅5⋅7⋅9+... till n terms.
My attempt:
I got the ith term to be (2i−1)(2i+1)(2i+3)(2i+5)
Expansion gives: 16i4+64i3+56i2+−16i−15
Required: n∑i=1(16i4+64i3+56i2+−16i−15)
Using summation identities, I got:
16n(n+1)(2n+1)(3n2+3n−1)30+64n2(n+1)24+56(n)(n+1)(2n+1)6−16n(n+1)2−15n
However, answer given is simply 110{(2n−1)(2n+1)(2n+3)(2n+5)(2n+7)+1⋅3⋅5⋅7}
No comments:
Post a Comment