For natural numbers m and n - if 1 - ym1 - yn - 1 - a1y - a2y2 - - - and a1 - a2 - 10- then m - n is

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Standard Formulae - 3

For natural n...

Question

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

$(35, 20)$

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$(45, 35)$

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$(35, 45)$

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$(20, 45)$

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Solution

The correct option is C
$(35, 45)$

Explanation for the correct answer:
Step1. Differentiate the given function:
${(1 - y)}^{m} {(1 + y)}^{n} = 1 + a_{1} y + a_{2} y^{2} + a_{3} y^{3} + \dots . . \Rightarrow m (- 1) {(1 - y)}^{m - 1} {(1 + y)}^{n} + {(1 - y)}^{m} (n) {(1 + y)}^{n - 1} = 0 + a_{1} + 2 a_{2} y + 3 a_{3} y^{2} + . . . [∵ \frac{d (u \times v)}{dx} = v \frac{du}{dx} + u \frac{dv}{dx} and \frac{d (x^{n})}{dx} = {nx}^{n - 1}] \Rightarrow - m {(1 - y)}^{m - 1} {(1 + y)}^{n} + {n (1 - y)}^{m} {(1 + y)}^{n - 1} = a_{1} + 2 a_{2} y + 3 a_{3} y^{2} + . . .____(1)$
Step2. Differentiate equation (1) again
$\begin{array}{rcl} - m (m - 1) (- 1) {(1 - y)}^{m - 2} {(1 + y)}^{n} + (- m) {(1 - y)}^{m - 1} (n) {(1 + y)}^{n - 1} + {n (m) (1 - y)}^{m - 1} {(1 + y)}^{n - 1} + (n) (n - 1) {(1 - y)}^{m} {(1 + y)}^{n - 2} & = & 0 + 2 a_{2} + 6 a_{3} y + . . . (2) \end{array}$
Step3.Simplify equation $(1) and (2)$
$P u t y = 0 in equation (1) - m + n = a_{1} [Given, a_{1} = a_{2} = 10] \Rightarrow - m + n = 10_____(3)$
$P u t y = 0 i n e q u a t i o n (2) m (m - 1) - m n - m n + n (n - 1) = 2 a_{2} \Rightarrow m^{2} - 2 m n + n^{2} - (m + n) = 2 a_{2} \Rightarrow {(n - m)}^{2} - (m + n) = 2 a_{2} \Rightarrow {(10)}^{2} - (m + n) = 2 (10) [Substitute the value of n - m = 10, a_{2} = 10] \Rightarrow m + n = 80______(4)$
Step 4. Calculate the value of $(m, n)$
Solve Equation $(3) a n d (4)$ we get
$n - m = 10 n + m = 80 \Rightarrow n = 45 and m = 35$
Hence, Option(C) is the correct answer.

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For natural numbers m and n , if (1–y)m(1+y)n=1+a1y+a2y2+…...anda1=a2=10, then (m,n) is

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