Binomial Expansion involving two terms? How would you find the 4th term in the expansion $(1+2x)^2 (1-6x)^{15}$?
Is there a simple way to do so?
Any help would be appreciated

$$(1+2x)^2 (1-6x)^{15}=\sum_{i=0}^{2}\binom{2}{i}(2x)^i\sum_{j=0}^{15}\binom{15}{j}(-6x)^j=$$
using $i+j=3$ for fourth term we get
$$\sum_{i+j=3}\binom{2}{i}2^i\binom{15}{j}(-6)^jx^3=$$
$$=\left(\binom{15}{3}(-6)^3+4\binom{15}{2}(-6)^2+4\binom{15}{1}(-6)^1\right)x^3=$$
$$=(-98280+15120-360)x^3=-83520x^3$$
