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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$-101376$

Vedclass · Answer

(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 $(a-2 b)^{12}$.

Similar Questions

The coefficient of $x^{91}$ in the series $^{100}C_1 \cdot 2^8 \cdot (1 - x)^{99} + ^{100}C_2 \cdot 2^7 \cdot (1 - x)^{98} + ^{100}C_3 \cdot 2^6 \cdot (1 - x)^{97} + \dots + ^{100}C_9 \cdot (1 - x)^{91}$ is equal to -

The coefficient of ${x^{39}}$ in the expansion of ${\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}$ is

If the second term of the expansion $\left[ a^{\frac{1}{13}} + \frac{a}{\sqrt{a^{-1}}} \right]^n$ is $14a^{5/2}$,then the value of $\frac{^nC_3}{^nC_2}$ is:

The terms containing $x^r y^s$ (for certain $r$ and $s$) are present in both the expansions of $(x+y^2)^{13}$ and $(x^2+y)^{14}$. If $\alpha$ is the number of such terms,then the sum $\alpha \sum_{r, s}(r+s) =$

The coefficient of $x^{70}$ in $x^2(1+x)^{98} + x^3(1+x)^{97} + x^4(1+x)^{96} + \ldots + x^{54}(1+x)^{46}$ is ${}^{99}C_p - {}^{46}C_q$. Then a possible value of $p+q$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module