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

Answer 1

akCn^(k-1)= [2^(k-1) +1]Cn^(k-1)

=2^(k-1)Cn^(k-1)+

2^(k-1)Cn^(k-1)的和是（1+2）^n的展开式,即3^n

Cn^(k-1)的和是2^n

所以原式＝3^n＋2^n