Problem: Let Sn​=1−2+3−4+…+(−1)n−1n,n=1,2,… Then S17​+S33​+S50​ equals:
Answer Choices:
A. 0
B. 1
C. 2
D. −1
E. −2
Solution:
Sn​=1−2+3−4+…+(−1)n−1n,n=1,2,3,…
For n even, Sn​=1+2+3+…+n−2(2+4+…+n)=21​n(n+1)−4⋅21​⋅2n​(1+2n​)
∴Sn​=21​n(n+1)−21​n(n+2)=−21​n
For n odd, Sn​=1+2+3+…+n−2(2+4+…+(n−1))
=21​n(n+1)−4⋅21​⋅2n−1​(1+2n−1​)
∴Sn​=21​n(n+1)−21​(n−1)(n+1)=21​(n+1)
∴S17​=9,S33​=17,S50​=−25∴S17​+S33​+S50​=1 .