Find sum to n terms of the series `1+(2+3)+(4+5+6)+"...."`.

Explanation

Now, number of terms in first bracket is `1`,in thr second bracket is 2, in the third bracket is 3, etc. Therefore, the number of terms in the nth bracket will be n.
Let the sum of the given sries of n terms =S
`therefore ` Number of terms in `S=1+2+3+"..."+n=(n(n+1))/(2)`
Also, the first term of S is 1 and common difference is also 1.
`therefore S= {(n(n+1))/(2)}/(2)[2*1+((n(n+1))/(2)-1)*1]`
`= (n(n+1))/(8)(4+n^(2)+n-2)`
`= (n(n+1)(n^(2)+n+2))/(8)`