If the coefficients of x and x^{2} in the exp | vedclass

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(B) The expansion is $(1+x)^{p}(1-x)^{q} = (1 + px + \frac{p(p-1)}{2}x^2 + \frac{p(p-1)(p-2)}{6}x^3 + \dots)(1 - qx + \frac{q(q-1)}{2}x^2 - \frac{q(q-

Vedclass · Accepted Answer

$23$

Vedclass · Answer

(B) The expansion is $(1+x)^{p}(1-x)^{q} = (1 + px + \frac{p(p-1)}{2}x^2 + \frac{p(p-1)(p-2)}{6}x^3 + \dots)(1 - qx + \frac{q(q-1)}{2}x^2 - \frac{q(q-1)(q-2)}{6}x^3 + \dots)$.The coefficient of $x$ is $p - q = -3$  $(1)$.The coefficient of $x^2$ is $\frac{p(p-1)}{2} - pq + \frac{q(q-1)}{2} = -5$.Multiplying by $2$: $p^2 - p - 2pq + q^2 - q = -10$.$(p-q)^2 - (p+q) = -10$.Since $p-q = -3$,we have $(-3)^2 - (p+q) = -10$,so $9 - (p+q) = -10$,which gives $p+q = 19$  $(2)$.Solving $(1)$ and $(2)$: $2p = 16 \implies p = 8$ and $q = 11$.The coefficient of $x^3$ is given by $\frac{p(p-1)(p-2)}{6} - \frac{p(p-1)}{2}q + p\frac{q(q-1)}{2} - \frac{q(q-1)(q-2)}{6}$.Substituting $p=8, q=11$:$= \frac{8 	imes 7 	imes 6}{6} - \frac{8 	imes 7}{2}(11) + 8 	imes \frac{11 	imes 10}{2} - \frac{11 	imes 10 	imes 9}{6}$.$= 56 - 308 + 440 - 165 = 23$.

If the coefficients of $x$ and $x^{2}$ in the expansion of $(1+x)^{p}(1-x)^{q}$,where $p, q \leq 15$,are $-3$ and $-5$ respectively,then the coefficient of $x^{3}$ is equal to $............$

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