`Sigma_(r=1)^(n)r(n-r)` बराबर है

Options

A. `1/6n(n-1)(2n+1)`
B. `[(n(n+1))/2]`
C. `(n^2(n+1))/6`
D. इनमे से कोई नहीं

Answer

D

Explanation

`underset(r=1)overset(n)(Sigma)r(n-r)=n underset(1)overset(n)(Sigma)" "r-underset(1)overset(n)(Sigma)r^2`
`=n(n(n+1))/2-(n(n+1)(2n+1))/6`
`(n(n+2))/6{3n-2-1}=(n(n+1)(n-1))/6`