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​=21​n(22​+2n+1​)
=21​n(2n+3​)=4n2+3n​