Question

Find a positive value of m for which the coefficient of x2 in expansion of (1+x)m is 6​

Answer 1

Step-by-step explanation:

6x^2(1+x)m= 0

6x^2 +mx+m=0

6x^2-3mx-2mx+m=0

-3x(-2x+1)+m(-2x+1) =0

(-3x+m)(-2x+1)=0

-3x+m=0. Or. - 2x+1=0

m=3x. - 2x=-1

x=1/2

m=3x

m=3*1/2

m=3/2

is the value for m

Answer 2

$\large \red{ \underline{ \underline{ \sf \: Given : \: \: }}}$

The coefficient of (x)² in expansion of $\sf{(1 + m)}^{2}$ is 6

$\large \red{ \underline{ \underline{ \sf \: Solution : \: \: }}}$

We know that , the general term of an expansion $\sf {(a + b)}^{n}$is

$\large \sf \fbox{T_{r + 1} = {}^{2}C_{r} \times {(a)}^{n - r} \times {(b)}^{r} }$

Suppose x² occurs in the $\sf {(r + 1)}^{th}$term of the expansion $\sf {(1 + x)}^{m}$

Therefore ,

$\to \sf T_{r + 1} = {}^{m}C_{r} \times {(1)}^{m - r} \times {(x)}^{r} \\ \\ \to \sf T_{r + 1} = {}^{m}C_{r} \times {(x)}^{r}$

Comparing the indices of x in (x)² and in $\sf T_{r + 1}$, we get r = 2 , so ,

$\sf \to T_{2+ 1} = {}^{m}C_{2} \times {(x)}^{2}$

it means ,

$\underline{ \: \: \sf \fbox{{}^{m}C_{2} = The \: coefficient \: of \: {(x)}^{2} } \: \: }$

$\sf \implies {}^{m}C_{2} = 6 \\ \\ \sf \implies \frac{m!}{(m - 2)!(2)! } = 6 \\ \\ \sf \implies \frac{m(m - 1)(m - 2)!}{(m - 2)! (2 \times 1)} = 6 \\ \\\sf \implies m \times(m - 1) = 12 \\ \\ \sf \implies {m}^{2} - m - 12 = 0 \\ \\\sf \implies {m}^{2} - 4m + 3m - 12 = 0 \\ \\\sf \implies m(m - 4) + 3(m - 4) = 0 \\ \\\sf \implies (m + 3)(m - 4) = 0 \\ \\\sf \implies m = - 3 \: \: or \: \: m = 4$

Hence , the required positive value of m is 4