For natural number m - n- if 1- y m 1- y n -1- a 1 y - a 2 y 2- and a 1- a 2-10- thenA- m nC- m - n -80D- m - n -20

The correct options are A m lt;n C m+n=80(1 y)^m(1+y)^n=(1 ^mC_1y+ ^mC_2y^2 ...)(1+ ^nC_1y+ ^nC_2y^2+...) =1+(n m)y+(m(m 1)2+n(n 1)2 mn)y^2+... Given, nbsp ...

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Standard XII

Mathematics

Middle Terms

For natural n...

Question

For natural number $m$ , $n$ , if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + . . .,$ and $a_{1} = a_{2} = 10$ , then

m < n

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m > n

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m + n = 80

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m - n = 20

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Solution

The correct options are
A $m < n$
C $m + n = 80$
$(1 - y)^{m} (1 + y)^{n} = (1 -^{m} C_{1} y +^{m} C_{2} y^{2} - . . .) (1 +^{n} C_{1} y +^{n} C_{2} y^{2} + . . .)$
$= 1 + (n - m) y + (\frac{m (m - 1)}{2} + \frac{n (n - 1)}{2} - m n) y^{2} + . . .$
Given,
$a_{1} = 10$
$\Rightarrow a_{1} = n - m = 10 \dots (1)$
$\Rightarrow a_{2} = \frac{m^{2} + n^{2} - m - n - 2 m n}{2} = 10$
$\Rightarrow (m - n)^{2} - (m + n) = 20$
$\Rightarrow m + n = 80 \dots (2)$
Solving $(1)$ and $(2)$ , we get $m = 35, n = 45$ .

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Middle Terms

Standard XII Mathematics

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For natural number m, n, if (1−y)m(1+y)n=1+a1y+a2y2+..., and a1=a2=10, then

The correct options are A m<n C m+n=80(1−y)m(1+y)n=(1− mC1y+ mC2y2−...)(1+ nC1y+ nC2y2+...) =1+(n−m)y+(m(m−1)2+n(n−1)2−mn)y2+... Given, a1=10 ⇒ a1=n−m=10⋯(1) ⇒ a2=m2+n2−m−n−2mn2=10 ⇒ (m−n)2−(m+n)=20 ⇒ m+n=80⋯(2) Solving (1) and (2), we get m=35, n=45.

For natural number $m$ , $n$ , if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + . . .,$ and $a_{1} = a_{2} = 10$ , then