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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

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

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HINT:

The $r$ th term $T_{r+1}$ of $(a+b)^n$ is $$\binom nr a^{n-r}b^r$$ where $0\le r\le n$

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