For natural numbers $m, n$ if $(1-y)^m(1+y)^n=1+a_1 y+a_2 y^2+\ldots$, and $a_1=a_2=10$ then $(\mathrm{m}, \mathrm{n})$ is

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Aicharya · Accepted Answer

$(1-y)^m(1+y)^n=\left[1-{ }^m C_1 y+{ }^m C_2 y^2-\ldotsight]\left[1+{ }^n C_1 y+{ }^n C_2 y^2+\ldotsight]$ $=1+(n-m)+\left\{\frac{m(m-1)}{2}+\frac{n(n-1)}{2}-m night\} y^2+\ldots .$. $	herefore a_1=n-m=10$ and $a_2=\frac{m^2+n-m-n-2 m n}{2}=10$ So, $n-m=10$ and $(m-n)^2-(m+n)=20 \quad \Rightarrow m+n=80$ $ 	herefore \mathrm{m}=35, \quad \mathrm{n}=45 $

For natural numbers $m, n$ if $(1-y)^m(1+y)^n=1+a_1 y+a_2 y^2+\ldots$, and $a_1=a_2=10$ then $(\mathrm{m},…

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