Problem: If xk+1=xk+21 for k=1,2,…,n−1 and x1=1, find x1+x2+…+xn.
Answer Choices:
A. 2n+1
B. 2n+3
C. 2n2−1
D. 4n2+n
E. 4n2+3n
Solution:
Since xk+1=xk+21 and x1=1,x2=x1+21=23,
x3=x2+21=24,x4=x3+21=25, and so forth with
xn=2n+1. The required sum is, therefore,
22+23+24+…+2n+1=21n(22+2n+1)
=21n(2n+3)=4n2+3n