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)=21n(n+1)−4⋅21⋅2n(1+2n)
∴Sn=21n(n+1)−21n(n+2)=−21n
For n odd, Sn=1+2+3+…+n−2(2+4+…+(n−1))
=21n(n+1)−4⋅21⋅2n−1(1+2n−1)
∴Sn=21n(n+1)−21(n−1)(n+1)=21(n+1)
∴S17=9,S33=17,S50=−25∴S17+S33+S50=1 .