Find the sum to n terms of the series :`1^2+(1^2+2^2)+(1^2+2^2+3^2)+ . . . `

Explanation

The sum of series `1^2+(1^2+2^2)+(1^2+2^2+3^2)+ldots n^2` can be calculated as

`=sum_(r=1)^(n) (1^2+2^2+ldots n^2)`

`=sum_(r=1)^(n)sum_(r=1)^(r) r^2`

`=sum_(r=1)^(n) (r(r+1)(2r+1))/6`

`=sum_(r=1)^(n) (2r^3+3r^2+r)/6`

`=1/3sum_(r=1)^(n) r^3+1/2 sum_(r=1)^(n) r^2+1/6sum_(r=1)^(n) r`

Since `sum n^2= (n(n+1))/2`

And `sum n^3= (n^2(n+1)^2)/4`

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