For the natural numbers m, n, if $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + \dots \dots + a_{m + n} y^{m + n}$ and $a_{1} = a_{2} = 10$ , then the value of (m + n) is equal to

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Detailed Solution

Given, $(1 - y)^{m} (1 + y)^{n} = 1 + a_{1} y + a_{2} y^{2} + \dots + a_{m + n} y^{m + n}$

Now, $(1 - y)^{m} (1 + y)^{n} = (1 - my +^{m} C_{2} y^{2} + \dots + (- 1)^{m} \cdot^{m} C_{m} y^{m}) \times (1 + ny +^{n} C_{2} y^{2} + . . +^{n} C_{n} y^{n})$

On expanding,

$(1 - y)^{m} (1 + y)^{n} = 1 + (n - m) y + (^{n} C_{2} - mn +^{m} C_{2}) y^{2} + \dots$

$∴ Coefficient of y = n - m$

Coefficient of $y^{2} =^{n} C_{2} - mn +^{m} C_{2}$

Given expression have

Coefficient of $y = a_{1} = 10$

Coefficient of $y^{2} = a_{2} = 10$

$∴ n - m = 10$ ...(i)

and $^{n} C_{2} +^{m} C_{2} - mn = 10$ ...(ii)

From Eq. (ii),

$\begin{array}{l} \frac{n!}{2! (n - 2)!} + \frac{m!}{2! (m - 2)!} - mn = 10 \\ \frac{n (n - 1)}{2} + \frac{m (m - 1)}{2} - mn = 10 \end{array}$

$\begin{array}{l} \Rightarrow n (n - 1) + m (m - 1) - 2 mn = 20 \\ \Rightarrow n^{2} - n + m^{2} - m - 2 mn = 20 \\ \Rightarrow (m^{2} + n^{2} - 2 mn) - (m + n) = 20 \\ \Rightarrow (m - n)^{2} - (m + n) = 20 \\ \Rightarrow (- 10)^{2} - (m + n) = 20 [using Eq . (i)] \\ \Rightarrow 100 - (m + n) = 20 \\ \Rightarrow m + n = 100 - 20 = 80 \end{array}$