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General Term of Binomial Expansion

If m and n ar...

Question

If m and n are the coefficients of $x^{a^{2}}$ and $x^{b^{2}}$ respectively in $(1 + x)^{a^{2} + b^{2}}$ , then

n = 2 m

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m + n = 0

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2 n = m

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m = n

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Solution

The correct option is D $m = n$
Given, $(1 + x)^{a^{2} + b^{2}}$

General term $T_{r + 1} =^{n} C_{r} x^{n - r} y^{r}$
$T_{r + 1} =^{a^{2} + b^{2}} C_{r} (1)^{a^{2} + b^{2} - r} (x)^{r}$

Given m is coefficient of $x^{a^{2}}$
$r = a^{2}$
$T_{r + 1} =^{a^{2} + b^{2}} C_{a^{2}} (1)^{a^{2} + b^{2} - a^{2}}$
$T_{a^{2} + 1} =^{a^{2} + b^{2}} C_{a^{2}} (1)^{b^{2}} (x)^{a^{2}}$

Given, n is coefficient of $x^{b^{2}}$
$r = b^{2}$
$T_{r + 1} =^{a^{2} + b^{2}} C_{b^{2}} (1)^{a^{2} + b^{2} - b^{2}} (x)^{b^{2}}$
$T_{r + 1} =^{a^{2} + b^{2}} C_{b^{2}} (1)^{a^{2}} (x)^{b^{2}}$
$^{a^{2} + b^{2}} C_{a^{2}} = m$
$^{a^{2} + b^{2}} C_{b^{2}} = n$

Using $^{n C_{r}} =^{n} C_{n - r}$
$n =^{a^{2} + b^{2}} C_{b^{2}} =^{a^{2} + b^{2}} C_{a^{2} + b^{2} - b^{2}}$
$=^{a^{2} + b^{2}} C_{a^{2}} = m$
so, $n = m$ .

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