If `sum_(r=1)^(n)r(r+1)(2r+3)=an^(4)+bn^(3)+cn^(2)+dn+e`, then.

Options

A. `a+c=b+d`
B. `e=0`
C. `a,b-2//3,c-1` are in A.P.
D. `c//a` is an integer

Answer

A, B, C, D

Explanation

`sum_(r=1)^(n)r(r+1)(2r+3)=sum_(r=1)^(n)2r^(3)+5r^(2)+3r`
`=2((n^(2)(n+1)^(2))/(4))+5((n(n+1)(2n+1))/(6))+3(n(n+1))/(2)`
`=(3n^(2)(n+1)^(2)+5n(n+1)(2n+1)+9n(n+1))/(6)impliesa=(3)/(6)=(1)/(2)`
`b=(16)/(6)=(8)/(3)`
`c=(27)/(6)=(9)/(2)`
`d=(14)/(6)=(7)/(3)`
`e=0`