Question

If the coefficient of ^2 is 6 in the expansion of (1 +x )^m then find the value of m .​

Answer 1

Appropriate Question:-

If m > 0 and coefficient of x² in the expansion of $\sf (1 + x) {}^{m}$is 6, then find the value of m.

Given:-

In expansion of $\sf (1 + x) {}^{m}$,

Coefficient of x² = 6.

To Find:-

• Value of m.

Solution:-

$\sf \large \because(a + b) {}^{n} ⇒T_{r + 1} = \: ^{n}C_{r}a {}^{n - r} b {}^{r}$

Here, we have

$\sf(1 + x) {}^{m}$⇒ n = m, a = 1, b = x.

$\sf \large \therefore T_{r + 1} = {}^{m}C _{r}a {}^{m - r} b {}^{r} = {}^{m} C_{r}(1) {}^{m - r} .(x) {}^{r}$

$\sf \large⇒T_{r + 1} = {}^{ m} C_rx {}^{r}$

For, coefficient of x² ; We have r = 2.

∴ By comparing; coefficient of x² is

$\sf {}^{m} C_2$

$\sf \large⇒ {}^{m} C_2 = 6$

$\sf \large⇒ \frac{m!}{(m - 2)!2!} = 6$

$\sf \large⇒ \frac{(m -2 )!((m - 1)m)}{(m - 2)!2!} = 6$

⇒ m(m – 1) = 6 × 2 = 12

⇒ m² – m – 12 = 0

⇒ m² – 4m + 3m – 12 = 0

⇒ m(m – 4) + 3(m – 4) = 0

⇒ m = – 3 (Or) m = 4

But m ≠ negative.

∴ m = 4.

Answer:-

${ \boxed{ \sf \huge \mathfrak \color{red} \therefore m = 4.}}$

Hope you have satisfied. ⚘