Question

Find the sum of the series till n terms `1*3*2^2 + 2*4*3^2 + 3*5*4^2...`?

Answer 1

Sum of the series is `1/5n^5+5/4n^4+7/2n^3+31/12n^2+79/30n+5/6`

Explanation:

It is evident that sum of the series tends to `oo`. However, let us find sum up to `n` terms.

`n^(th)` term of the series `1*3*2^2+2*4*3^2+3*5*4^2+....` is

`n(n+2)(n+1)^2` and we have to find `sum_1^n(n(n+2)(n+1)^2)`

As `n(n+2)(n+1)^2`

= `(n^2+2n)(n^2+2n+1)`

= `n^4+2n^3+n^2+2n^3+4n^2+2n`

= `n^4+3n^3+5n^2+2n`

Hence `sum_1^n(n^4+3n^3+5n^2+2n)`

= `sum_1^n n^4+3sum_1^n n^3+5sum_1^n n^2+2sum_1^n n`

= `(n(n+1)(2n+1)(3n^2+3n-1))/30+3(n^2(n+1)^2)/4+5(n(n+1)(2n+1))/6+2(n(n+1))/2`

= `1/30(n^2+n)(2n+1)(3n^2+3n-1)+3/4n^2(n+1)^2+5/6n(n+1)(2n+1)+n(n+1)`

= `1/30(2n^3+3n^2+n)(3n^2+3n-1)+3/4n^2(n^2+2n+1)+5/6(n^2+1)(2n+1)+n(n+1)`

= `1/30(6n^5+15n^4+10n^3-n)+3/4(n^4+2n^3+n^2)+5/6(2n^3+n^2+2n+1)+n^2+n`

= `1/5n^5+n^4(1/2+3/4)+n^3(1/3+3/2+5/3)+n^2(3/4+5/6+1)+n(-1/30+5/3+1)+5/6`

= `1/5n^5+5/4n^4+7/2n^3+31/12n^2+79/30n+5/6`