Find a positive value of m for which the coef | vedclass

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

Question

(A) The general term $(T_{r+1})$ in the binomial expansion of $(1+x)^{m}$ is given by $T_{r+1} = {}^{m}C_{r} (1)^{m-r} (x)^{r} = {}^{m}C_{r} x^{r}$.Fo

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The general term $(T_{r+1})$ in the binomial expansion of $(1+x)^{m}$ is given by $T_{r+1} = {}^{m}C_{r} (1)^{m-r} (x)^{r} = {}^{m}C_{r} x^{r}$.For the coefficient of $x^{2}$,we set $r=2$.Thus,the coefficient of $x^{2}$ is ${}^{m}C_{2}$.Given that the coefficient is $6$,we have ${}^{m}C_{2} = 6$.$\frac{m(m-1)}{2} = 6$$m(m-1) = 12$$m^{2} - m - 12 = 0$$(m-4)(m+3) = 0$Since $m$ must be positive,$m=4$.

Find a positive value of $m$ for which the coefficient of $x^{2}$ in the expansion $(1+x)^{m}$ is $6$.

Similar Questions

If $x$ is numerically so small that $x^2$ and higher powers of $x$ can be neglected, then $\left(1+\frac{2x}{3}\right)^{3/2} \cdot (32+5x)^{-1/5}$ is approximately equal to

The expansion of $(1+x+x^2)^{-3/2}$ in powers of $x$ is valid if

If $x$ is small,so that $x^2$ and higher powers can be neglected,then the approximate value for $\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}$ is

The coefficient of $x^2$ in the expansion of $(1+x)^2(8-x)^{-\frac{1}{3}}$ is

If $(2-5x)^{-1/5} = a_0 + a_1x + a_2x^2 + \ldots$,then $\frac{a_1}{a_2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module