If in the expansion of , the coefficients of x and are 1 and –2, respectively, then is equal to : [2025]

Question

If in the expansion of ${(1 + x)}^{p} {(1 - x)}^{q}$ , the coefficients of x and $x^{2}$ are 1 and –2, respectively, then $p^{2} + q^{2}$ is equal to : [2025]

Smart Achievers · Accepted Answer

(4)

${(1 + x)}^{p} {(1 - x)}^{q} = ({}^{p}C_{0} + C_{1} x + {}^{p}C_{2} x^{2} + . . .) ({}^{q}C_{0} - {}^{q}C_{1} x + {}^{p}C_{2} x^{2} + . . .)$

Coeff. of $x \equiv - {}^{p}C_{0} {}^{q}C_{1} + {}^{p}C_{1} {}^{q}C_{0} = 1$

$\Rightarrow p - q = 1$ ... (i)

Coeff. of $x^{2} \equiv {}^{p}C_{0} {}^{q}C_{2} - {}^{p}C_{1} {}^{q}C_{1} + {}^{p}C_{2} {}^{q}C_{0} = - 2$

$\Rightarrow \frac{q (q - 1)}{2} - p q + \frac{p (p - 1)}{2} = - 2$

$\Rightarrow q^{2} - q - 2 p q + p^{2} - p = - 4$

$\Rightarrow 1 - (p + q) = - 4$ $[∵ q^{2} + p^{2} - 2 p q = {(p - q)}^{2}]$

$\Rightarrow p + q = 5$ ... (ii)

On solving equation (i) and (ii), we get p = 3 and q = 2.

So, . $p^{2} + q^{2} = 13$

Solution

Q.
If in the expansion of ${(1 + x)}^{p} {(1 - x)}^{q}$ , the coefficients of x and $x^{2}$ are 1 and –2, respectively, then $p^{2} + q^{2}$ is equal to : [2025]

1 18

2 8

3 20

4 13

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