Problem: Let (1+x+x2)n=a0​+a1​x+a2​x2+⋯+a2n​x2n 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​−a1​x+a2​x2−a3​x3+…(1+x+x2)n≡a0​+a1​x+a2​x2+a3​x3+…
∴2(a0​+a2​x2+a4​x4+…+a2n​x2n)≡(1−x+x2)n+(1+x+x2)n
Let x=1∴2(a0​+a2​+…+a2n​)=1n+3n∴2s=1n+3n∴s=23n+1​