Question

Find the coefficient of x^3 in the expansion of (1-3x)(1+2x)^6
Please Help! Thank you!

Answer 1

\[(1+2x)^6=\sum_{i=0}^6\binom{6}{i}(2x)^i\\
(1-3x)(1+2x)^6=(1-3x)\left[\sum_{i=0}^6\binom{6}{i}2^ix^i\right]=\sum_{i=0}^6\binom{6}{i}2^ix^i-\sum_{k=0}^63\binom{6}{k}2^kx^{k+1}\]
To get an x^3, you need i=3 and k=2.

Answer 2

Thus, the coefficient of x^3 is
\[\binom{6}{3}2^3-3\binom{6}{2}2^2\]