已知数列{an}的通项公式为an=2^(n-1)+1则a1Cn^0+a2Cn^1+a3Cn^2+...+a[n+1]Cn^n?a1×Cn0+a2×Cn1……+a(n+1)×Cnn=(2^0+1)*C(n,0)+(2^1+1)C(n,1)+(2^2+1)C(n,2)+.+(2^n+1)*C(n,n)=[2^0*C(n,0)+2^1*C(n,1)+2^2*C(n,2)+.+2^n*C(n,n)]+[C(n,0)+C(n,1)+C(n

Question

已知数列{an}的通项公式为an=2^(n-1) +1 则a1Cn^0 +a2Cn^1+a3Cn^2+...+a[n+1]Cn^n?
a1×Cn0+a2×Cn1……+a(n+1)×Cnn
=(2^0+1)*C(n,0)+(2^1+1)C(n,1)+(2^2+1)C(n,2)+.+(2^n +1)*C(n,n)
=[2^0*C(n,0)+2^1*C(n,1)+2^2*C(n,2)+.+2^n *C(n,n)]+[C(n,0)+C(n,1)+C(n,2)+.+C(n,n)]
=(1+2)^n+2^n
=3^n+2^n
此过程中的倒数第二步如何解释?

Answer 1

就是二项式定理的逆用
(1+2)^n=2^0*C(n,0)+2^1*C(n,1)+2^2*C(n,2)+.+2^n *C(n,n)]
2^n=(1+1)^n=C(n,0)+C(n,1)+C(n,2)+.+C(n,n)]


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