Question

The question is below?

For the series `S=1+1/(1+3)(1+2)^2+1/(1+3+5)*(1+2+3)^2+1/(1+3+5+7)(1+2+3+4)^2+....` What is the sum to nth term?

Answer 1

`S_n=n/24{2n^2+9n+13}`

Explanation:

Here ,

`S=1^2/1+(1+2)^2/(1+3)+(1+2+3)^2/(1+3+5)+(1+2+3+4)^2/(1+3+5+7)+...`

Let , `r^(th)term =t_r ,then`

`t_r=(1+2+3+...+r)^2/(1+3+5+...+(2r-1))=(sumr)^2/(sum(2r-1))`

`:.t_r=[r/2(r+1)]^2/(2sumr-sum1)=[r^2/4(r+1)^2]/[2(r/2(r+1))-r)=[r^2/4(r+1)^2]/(r(r+1-1)`

`:.t_r=[r^2/4(r+1)^2]/r^2=1/4(r^2+2r+1)`

So,

`S_n=sum_(r=1)^n1/4[r^2+2r+1]=1/4{sum_(r=1)^nr^2+2sum_(r=1)^nr+sum_(r=1)^n1}`

`:.S_n=1/4{n/6(n+1)(2n+1)+2*n/2(n+1)+n}`

`:.S_n=1/4xxn/6{(n+1)(2n+1)+6(n+1)+6}`

`=>S_n=n/24{2n^2+n+2n+1+6n+6+6}`

`=>S_n=n/24{2n^2+9n+13}`

Answer 2

`=(2N^3+9N^2+13N)/24`

Explanation:

If you look at `n`th term, it is the square of the sum of the first `n` integers divided by the sum of the first `n` odd integers.

Recall that the sum of the first `n` integers is `1/2n(n+1)`. Squaring this gives `1/4n^2(n+1)^2`.

Now, the sum of the first `n` odd integers is `n^2`.

Thus, we have each term being `1/4(n+1)^2`. Sum this:

`sum_(n=1)^N1/4(n+1)^2`

`=1/4sum_(n=1)^N(n+1)^2`

`=1/4(sum_(n=1)^(N+1)n^2-1)`

Recall the formula for the sum of the first `n` perfect squares:

`=1/4((2(N+1)^3+3(N+1)^2+(N+1))/6-1)`

`=(2N^3+9N^2+13N)/24`