已知数列{an}的通项公式是an{2n-1(n为奇数)3^n(n为偶数),求数列an的前2n项和Sn

Answer 1

2n项的和式里肯定有n项是奇数,n项是偶数,所以分开求和即可.Sum是求和符号.
S奇数 = Sum (k=1,3,5,...,2n-1) (2k-1)
= 1 + 5 + 9 + ...+ (4n-3) = (4n-2)*n/2；
S偶数 = Sum (k = 2,4,6,...,2n) 3^k
= 9 + 81 + 81*9 + ...+ 9^n
= 9(9^n - 1)/8,于是
S2n = S奇数 + S偶数 = (4n-2)*n/2 + 9(9^n - 1)/8 = 2n^2 - n + 9(9^n - 1)/8.