Problem: The third term in the expansion of (x​a​−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=x​a​ and s=−a2x​​ in the third term, 15(x​a​)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​(x​a​)4(−a2x​​)2=x15​