Q. 18

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

Answer :

The general term T_{r+1} in the binomial expansion is given by T_{r+1} = ^{n}C_{r} a^{n-r} b^{r}

Here a = 1, b = x and n = m

Putting the value

T_{r+1} = ^{m}C_{r} 1^{m-r} x^{r}

= ^{m}C_{r} x^{r}

We need coefficient of x^{2}

∴ putting r = 2

T_{2+1} = ^{m}C_{2} x^{2}

The coefficient of x^{2} = ^{m}C_{2}

Given that coefficient of x^{2} = ^{m}C_{2} = 6

⇒

⇒

⇒ m(m-1) = 12

⇒ m^{2}- m - 12 =0

⇒ m^{2}- 4m +3m - 12 =0

⇒ m(m-4) + 3(m-4) = 0

⇒ (m+3)(m - 4)= 0

⇒ m = - 3, 4

we need positive value of m so m = 4

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