Problem: 2−(2k+1)−2−(2k−1)+2−2k2^{-(2 k+1)}-2^{-(2 k-1)}+2^{-2 k}2−(2k+1)−2−(2k−1)+2−2k is equal to
Answer Choices:
A. 2−2k2^{-2 k}2−2k
B. 2−(2k−1)2^{-(2 k-1)}2−(2k−1)
C. −2−(2k+1)-2^{-(2 k+1)}−2−(2k+1)
D. 000
E. 222 Solution:
The given expression =2−2k(2−1−2+1)=−2−2k/2=−2−(2k+1)=2^{-2 k}\left(2^{-1}-2+1\right)=-2^{-2 k} / 2=-2^{-(2 k+1)}=2−2k(2−1−2+1)=−2−2k/2=−2−(2k+1) \quad