Problem: The third term in the expansion of (xa−a2x)6 is, when simplified:
Answer Choices:
A. x15
B. −x15
C. −a96x2
D. a320
E. −a320
Solution:
Since (r+s)4=r6+6r5s+15r4s2+⋯, we have, letting r=xa and s=−a2x in the third term, 15(xa)4(−a2x)2=x15
or
Use the formula for the general term of (r+s)n,n positive integer: (k+1)th term
=1⋅2⋯kn(n−1)⋯(n−k+1)(r)n−k(s)k
For k=2 we have 1⋅26⋅5(xa)4(−a2x)2=x15