化简:4n+3×4n-1+32×4n-2+…+3n-1×4+3n.

问题描述:

化简:4n+3×4n-1+32×4n-2+…+3n-1×4+3n

设Sn=4n+3×4n-1+32×4n-2+…+3n-1×4+3n

4
3
Sn
4n+1
3
+4n+3×4n−1+32×4n−2+…+3n-2×42+3n-1×4,
两式相减,得
1
3
Sn
=
1
3
×4n+13n

Sn4n+13n+1
答案解析:利用错位相减法求解.
考试点:数列的求和.
知识点:本题考查数列的前n项和的求法,解题时要认真审题,注意错位相减法的合理运用.