Question

已知数列{an}的通项公式为an=2^(n-1)+1,则a1×Cn0+a2×Cn1……+a(n+1)×Cnn=？

Answer 1

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

Answer 2

a1×Cn0+a2×Cn1……+a(n+1)×Cnn
＝(2^0+1)×Cn0+(2^1+1) ×Cn1+(2^2+1)×Cn2+……+(2^n+1)×Cnn
＝[1^n×2^0×Cn0 + 1^(n-1)×2^1×Cn1 + 1^(n-2)×2^2×Cn2 +……+ 1^0×2^n×Cnn]
+(Cn0+Cn1+Cn2+……+Cnn)
＝(1+2)^n+(1+1)^n
＝3^n+2^n

Answer 3

将其拆分即=（a1+a2+a3+a4+....+an）+（Cn0+Cn1+Cn2+....+Cnn）。这样求解很简单！