Find the sum `sum_(r=1)^(n) r^(2) = (""^(n)C_(r))/(""^(n)C_(r-1))`

Answer

`(n(n+1)(n+2))/(6)`

Explanation

`t_(r)=r^(2)(.^(n)C_(r))/(.^(n)C_(r)-1)`
`=r^(2)(n-r+1)/(r)`
`=r(n+1-r)`
`=(n+1)r-r^(2)`
`:.` Sum `= (n+1)underset(r=1)overset(n)sumr-underset(r=1)overset(n)sumr^(2)`
`=(n+1)(n(n+1))/(2)-(n(n+1)(2n+1))/(6)`
`= (n(n+1))/(6){3(n+1) - (2n+1)}`
`= (n(n+1)(n+2))/(6)`