# Find the coefficient of $a^5b^7$ in $(a-2b)^{12}$

$\begin{array}{1 1} -101376 \\ 69341 \\ 45623\\ 62013\end{array}$

• $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n, where b^0 = 1 = a^{n-n} • Looking at the pattern of the successive terms we can say that the (r + 1)\text{th} term T_{r+1} = ^n\large C$$_r\;a^{n–r}b^r$. This is also called the general term of the expansion.
In the expression $(a-2b)^{12}$, assume that $a^5b^7$ occurs in the $(r+1)\text{th}$ term.
$\Rightarrow T_{r+1} = \; ^{12} \large C$$_r\;a^{12–r}(-2b)^r. Comparing the indices of a and b, we can see that 12 - r = 5 \rightarrow r = 7 \Rightarrow Coefficient of a^5b^7 = \;^{12} \large C$$_7 \; a^{12-7}(-2b)^{7}$