Find the coefficient of a^{5} b^{7} in (a-2 b | vedclass

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(B) The general term $(T_{r+1})$ in the binomial expansion of $(a+x)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} x^{r}$.For the expansion $(a-2b)^

Vedclass · Accepted Answer

$-101376$

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(B) The general term $(T_{r+1})$ in the binomial expansion of $(a+x)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} x^{r}$.For the expansion $(a-2b)^{12}$,the general term is:$T_{r+1} = {}^{12}C_{r} (a)^{12-r} (-2b)^{r} = {}^{12}C_{r} (-2)^{r} a^{12-r} b^{r}$.To find the coefficient of $a^{5} b^{7}$,we compare the powers of $b$:$r = 7$.Substituting $r = 7$ into the expression for the coefficient:Coefficient $= {}^{12}C_{7} (-2)^{7}$.Calculating the values:${}^{12}C_{7} = \frac{12!}{7!5!} = \frac{12 	imes 11 	imes 10 	imes 9 	imes 8}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 792$.$(-2)^{7} = -128$.Coefficient $= 792 	imes (-128) = -101376$.

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

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