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Answer:
∑nr(r+1
(2r+3)=r=1∑n(2r3+3r2+2r2+3r)=r=1∑n(2r3+5r2+3r)
=2(2n(n+1))2+5(6n(n+1)(2n+1))+3(2n(n+1))
=21(n2(n2+1+2n))+65(n(2n2+3n+1))+23(n2+n)
=21(n4+2n3+n2)+65(2n3+3n2+n)+23(n2+n)
=21n4+616n3+29n2+614n
BRO HENCE IS NOT VERTIFIED
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