If in the expansion of (1+x)^p(1-x)^q,the coe | vedclass

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(C) The expansion is given by $(1+x)^p(1-x)^q = (1 + px + \frac{p(p-1)}{2}x^2 + \dots)(1 - qx + \frac{q(q-1)}{2}x^2 - \dots)$.The coefficient of $x$ i

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) The expansion is given by $(1+x)^p(1-x)^q = (1 + px + \frac{p(p-1)}{2}x^2 + \dots)(1 - qx + \frac{q(q-1)}{2}x^2 - \dots)$.The coefficient of $x$ is $p - q = 1$.The coefficient of $x^2$ is $\frac{q(q-1)}{2} - pq + \frac{p(p-1)}{2} = -2$.Multiplying by $2$,we get $q^2 - q - 2pq + p^2 - p = -4$.Rearranging,$(p^2 - 2pq + q^2) - (p + q) = -4$.$(p - q)^2 - (p + q) = -4$.Since $p - q = 1$,we have $1^2 - (p + q) = -4$,which implies $p + q = 5$.Solving $p - q = 1$ and $p + q = 5$,we get $2p = 6 \implies p = 3$ and $q = 2$.Thus,$p^2 + q^2 = 3^2 + 2^2 = 9 + 4 = 13$.

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 :

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