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.
答
设Sn=4n+3×4n-1+32×4n-2+…+3n-1×4+3n,
则
Sn=4 3
+4n+3×4n−1+32×4n−2+…+3n-2×42+3n-1×4,4n+1 3
两式相减,得
Sn=1 3
×4n+1−3n,1 3
Sn=4n+1−3n+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.