Problem: Let (1+x+x2)n=a0+a1x+a2x2+⋯+a2nx2n be an identity in x. If we let s=a0+a2+a4+⋯+a2n, then s equals:
Answer Choices:
A. 2n
B. 2n+1
C. 23n−1
D. 23n
E. 23n+1
Solution:
(1−x+x2)n≡a0−a1x+a2x2−a3x3+…(1+x+x2)n≡a0+a1x+a2x2+a3x3+…
∴2(a0+a2x2+a4x4+…+a2nx2n)≡(1−x+x2)n+(1+x+x2)n
Let x=1∴2(a0+a2+…+a2n)=1n+3n∴2s=1n+3n∴s=23n+1