Problem: When (a−b)n,n≥2,abî€ =0, is expanded by the binomial theorem, it is found that, when a=kb, where k is a positive integer, the sum of the second and third terms is zero. Then n equals:
Answer Choices:
A. 21​k(k−1)
B. 21​k(k+1)
C. 2k−1
D. 2 k
E. 2k+1
Solution:
(a−b)n=an−nan−1b+1⋅2n(n−1)​an−2b2−…
∴−n(kb)n−1b+2n(n−1)​(kb)n−2b2=0
∴−kn−1+2n−1​kn−2=0.∴n−1=kn−22kn−1​∴n=2k+1.