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

Vedclass pdf generator app on play store
Vedclass iOS app on app store

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$

Similar Questions

The ratio of the coefficient of terms ${x^{n - r}}{a^r}$and ${x^r}{a^{n - r}}$ in the binomial expansion of ${(x + a)^n}$ will be

If the number of integral terms in the expansion of $\left(3^{\frac{1}{2}}+5^{\frac{1}{8}}\right)^{\text {n }}$ is exactly $33,$ then the least value of $n$ is

  • [JEE MAIN 2020]

If coefficients of ${(2r + 1)^{th}}$ term and ${(r + 2)^{th}}$ term are equal in the expansion of ${(1 + x)^{43}},$ then the value of $r$ will be

The total number or irrational terms in the binomial expansion of $\left( {{7^{1/5}} - {3^{1/10}}} \right)^{60}$ is 

  • [JEE MAIN 2019]

The term independent of $x$ in ${\left( {\sqrt x - \frac{2}{x}} \right)^{18}}$ is