For natural numbers m, n,if (1 - y)^m(1 + y)^ | vedclass

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(D) Given $(1 - y)^m(1 + y)^n = 1 + a_1y + a_2y^2 + \ldots$Expanding the binomials:$(1 - my + \frac{m(m-1)}{2}y^2 - \ldots)(1 + ny + \frac{n(n-1)}{2}y

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$(35, 45)$

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(D) Given $(1 - y)^m(1 + y)^n = 1 + a_1y + a_2y^2 + \ldots$Expanding the binomials:$(1 - my + \frac{m(m-1)}{2}y^2 - \ldots)(1 + ny + \frac{n(n-1)}{2}y^2 + \ldots) = 1 + a_1y + a_2y^2 + \ldots$Comparing the coefficient of $y$:$a_1 = n - m = 10 \implies n = m + 10$Comparing the coefficient of $y^2$:$a_2 = \frac{n(n-1)}{2} - nm + \frac{m(m-1)}{2} = 10$$n^2 - n - 2nm + m^2 - m = 20$$(n - m)^2 - (n + m) = 20$Since $n - m = 10$,we have $(10)^2 - (m + 10 + m) = 20$$100 - 2m - 10 = 20$$90 - 2m = 20$$2m = 70 \implies m = 35$$n = 35 + 10 = 45$Thus,$(m, n) = (35, 45)$.

For natural numbers $m, n$,if $(1 - y)^m(1 + y)^n = 1 + a_1y + a_2y^2 + \ldots$ and $a_1 = a_2 = 10$,then $(m, n) = \_\_\_\_\_\_$.

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