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If Sigma(r=1)^(n) r(r+1)(2r +3)=an^4+bn...

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

A

a-b=d-c

B

e=0

C

`a,b-2//3,c-1 " are in " in A.P`

D

`(b+d)//a` is an integer

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The correct Answer is:
A, B, C, D
`an^(4)+bn^(3)+cn^(3)+dn+e`
`=2sum_(r=1)^(n)r(r+1)(r+2)-sum_(r=1)^(n)r(r+1)`
`=2/4n(n+1)(n+2)(n+3)-1/3n(n+1)(n+2)`
`=1/6(3n^(4)+16n^(3)+27n^(2)+14n)`
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