Find the cocfficient of $a^{5} b^{7}$ in $(a-2 b)^{12}$
It is known that $(r+1)^{\text {th }}$ term, $\left(T_{r+1}\right),$ in the binomial expansion of $(a+b)^{n}$ is given by
$T_{n+1}=^{n} C_{r} a^{n-r} b^{r}$
Assuming that $a^{5} b^{7}$ occurs in the $(r+1)^{th}$ term of the expansion $(a-2 b)^{12},$ we obtain
${T_{r + 1}} = {\,^{12}}{C_r}{(a)^{12 - r}}{( - 2b)^r} = {\,^{12}}{C_r}{( - 2)^r}{(a)^{12 - r}}{(b)^r}$
Comparing the indices of a and $b$ in $a^{5} b^{7}$ in $T_{r+1},$
We obtain $r=7$
Thus, the coefficient of $a^{5} b^{7}$ is
${\,^{12}}{C_7}{( - 2)^7} = \frac{{12!}}{{7!5!}} \cdot {2^7} = \frac{{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!}}{{5 \cdot 4 \cdot 3 \cdot 2 \cdot 7!}} \cdot {( - 2)^7}$
$ = - (792)(128) = - 101376$
The value of $x$ in the expression ${[x + {x^{{{\log }_{10}}}}^{(x)}]^5}$, if the third term in the expansion is $10,00,000$
If sum of the coefficient of the first, second and third terms of the expansion of ${\left( {{x^2} + \frac{1}{x}} \right)^m}$ is $46$, then the coefficient of the term that doesnot contain $x$ is :-
The sum of all those terms which are rational numbers in the expansion of $\left(2^{1 / 3}+3^{1 / 4}\right)^{12}$ is:
The coefficient of $x^9$ in the expansion of $(1+x)\left(1+x^2\right)\left(1+x^3\right) \ldots . .\left(1+x^{100}\right)$ is