Problem: When (aβb)n,nβ₯2,abξ =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. 21βk(kβ1)
B. 21βk(k+1)
C. 2kβ1
D. 2 k
E. 2k+1
Solution:
(aβb)n=anβnanβ1b+1β
2n(nβ1)βanβ2b2ββ¦
β΄βn(kb)nβ1b+2n(nβ1)β(kb)nβ2b2=0
β΄βknβ1+2nβ1βknβ2=0.β΄nβ1=knβ22knβ1ββ΄n=2k+1.