Find the sum of the series `1xxn+2(n-1)+3xx(n-2)++(n-1)xx2+nxx1.`

Explanation

Let `T_(r)` be the rth term of the given series. Then,
`T_(r)=rxx{n-(r-1)}`
=r(n-r+1)
=r{(n+1)-r}
=(n+1)r-`r^(2)`
`thereforesum_(r=1)^(n)T_(r)=sum_(r=1)^(n)[(n+1)r-r^(2)]`
`=(n+1)(sum_(r=1)^(n)r)-(sum_(r=1)^(n)r^(2))`
`=(n+1)(n(n+1))/2-(n(n+1)(2n+1))/6`
`=(n(n+1)(n+2))/6`