求证:Cn0+2Cn1+3Cn2+…+(n+1)Cnn=2n+n2n-1

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求证:Cn0+2Cn1+3Cn2+…+(n+1)Cnn=2n+n2n-1求证:Cn0+2Cn1+3Cn2+…+(n+1)Cnn=2n+n2n-1求证:Cn0+2Cn1+3Cn2+…+(n+1)Cnn=

求证:Cn0+2Cn1+3Cn2+…+(n+1)Cnn=2n+n2n-1
求证:Cn0+2Cn1+3Cn2+…+(n+1)Cnn=2n+n2n-1

求证:Cn0+2Cn1+3Cn2+…+(n+1)Cnn=2n+n2n-1
已知Cni=Cn(n-i)
则原等式左边=
Cnn+2Cn(n-1)+3Cn(n-2)+…+(n+1)Cn0
两式相加得
2[Cn0+2Cn1+3Cn2+…+(n+1)Cnn]
=(n+2)(Cn0+Cn1+…Cnn)
=(n+2)2^n

Cn0+2Cn1+3Cn2+…+(n+1)Cnn
=(n+2)2^(n-1)
=2^n+n2^(n-1)