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Question Number 63831 by gunawan last updated on 10/Jul/19

If C_r be the coefficient of x^r in (1+x)^n , then the value of Σ_(r=0) ^n (r+1)^2 C_r is

IfCrbethecoefficientofxrin(1+x)n,thenthevalueof∑nr=0(r+1)2Cris

Answered by mr W last updated on 10/Jul/19

Σ_(r=0) ^n C_r x^r =(1+x)^n Σ_(r=0) ^n C_r x^(r+1) =(1+x)^n x Σ_(r=0) ^n (r+1)C_r x^r =(1+x)^n +n(1+x)^(n−1) x Σ_(r=0) ^n (r+1)C_r x^(r+1) =(1+x)^n x+n(1+x)^(n−1) x^2 Σ_(r=0) ^n (r+1)^2 C_r x^r =(1+x)^n +n(1+x)^(n−1) x+2n(1+x)^(n−1) x+n(n−1)(1+x)^(n−2) x^2 let x=1 Σ_(r=0) ^n (r+1)^2 C_r =2^n +n2^(n−1) +2n2^(n−1) +n(n−1)2^(n−2) =(n+1)2^n +n(n+1)2^(n−2) =(n+1)(n+4)2^(n−2)

∑nr=0Crxr=(1+x)n∑nr=0Crxr+1=(1+x)nx∑nr=0(r+1)Crxr=(1+x)n+n(1+x)n−1x∑nr=0(r+1)Crxr+1=(1+x)nx+n(1+x)n−1x2∑nr=0(r+1)2Crxr=(1+x)n+n(1+x)n−1x+2n(1+x)n−1x+n(n−1)(1+x)n−2x2letx=1∑nr=0(r+1)2Cr=2n+n2n−1+2n2n−1+n(n−1)2n−2=(n+1)2n+n(n+1)2n−2=(n+1)(n+4)2n−2

Commented by gunawan last updated on 10/Jul/19

wow thank you Sir

wowthankyouSir

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