For natural numbers m and n (1−y)m(1+y)n=1+a1y+a2y2+… and a1=a2=10, then (m,n) equals

(1−y)m(1+y)n=1−my+mC2y2−…(1+ny+nC2y2+… =1+(n−m)y+ nC2+mC2−mny2+……We are given n-m=a1=10and nC2+mC2−mn=a2=10⇒ n=m+10 and 12n(n−1)+12m(m−1)−mn=10⇒ (m+10)(m+9)+m(m−1)−2m(m+10)=20⇒ m2+19m+90+m2−m−2m2−20m=20⇒ −2m=−70⇒m=35.∴ n=45

For natural numbers m and n (1−y)m(1+y)n=1+a1y+a2y2+… and a1=a2=10, then (m,n) equals

(1−y)m(1+y)n=1−my+mC2y2−…(1+ny+nC2y2+… =1+(n−m)y+ nC2+mC2−mny2+……We are given n-m=a1=10and nC2+mC2−mn=a2=10⇒ n=m+10 and 12n(n−1)+12m(m−1)−mn=10⇒ (m+10)(m+9)+m(m−1)−2m(m+10)=20⇒ m2+19m+90+m2−m−2m2−20m=20⇒ −2m=−70⇒m=35.∴ n=45

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For natural numbers $m$ and $n$ $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + \dots$ and $a_{1} = a_{2} = 10,$ then $(m, n)$ equals

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$(35, 45)$

$(20, 45)$

$(35, 20)$

$(45, 35)$

answer is A.

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Detailed Solution

$\begin{array}{r} (1 - y)^{m} (1 + y)^{n} = (1 - m y +^{m} C_{2} y^{2} - \dots) (1 + n y +^{n} C_{2} y^{2} + \dots) \end{array}$
$= 1 + (n - m) y + (^{n} C_{2} +^{m} C_{2} - m n) y^{2} + \dots$ …
We are given
$n - m = a_{1} = 10$
and $^{n} C_{2} +^{m} C_{2} - m n = a_{2} = 10$
$\Rightarrow n = m + 10$ and
$\begin{array}{ll} \frac{1}{2} n (n - 1) + \frac{1}{2} m (m - 1) - m n = 10 \\ \Rightarrow (m + 10) (m + 9) + m (m - 1) - 2 m (m + 10) = 20 \\ \Rightarrow m^{2} + 19 m + 90 + m^{2} - m - 2 m^{2} - 20 m = 20 \\ \Rightarrow - 2 m = - 70 \Rightarrow m = 35 . \\ ∴ n = 45 \end{array}$