Problem: When (1−a1)6 is expanded the sum of the last three coefficients is:
Answer Choices:
A. 22
B. 11
C. 10
D. −10
E. −11
Solution:
The last three terms of the expansion of (1−a1)6 correspond in inverse order to the first three terms in the expansion of (−a1+1)6.
These are 1−6⋅a1+15(−a1)2, and the sum of these coefficients is 1−6+15=10
or
Expanding completely we have 1−a6+a215−a320+a415−a56+a61;15−6+1=10