"For the natural numbers m, n, if ${(1 - y)^m}{(1 + y)^n} = 1 + {a_1}y + {a_2}{y^2} + .... + {a_{m + n}}{y^{m + n}}$ and ${a_1} = {a_2} = 10$, then the value of (m + n) is equal to :"

Question

Accepted Answer

"${(1 - y)^m}{(1 + y)^n} = 1 + {a_1}y + {a_2}{y^2} + .... + {a_{m + n}}{y^{m + n}}$

Given, (${a_1} = {a_2} = 10$)$$(1 - my + {}^m{C_2}{y^2} + .....)(1 + ny + {}^n{C_2}{y^2} + .....) = 1 + {a_1}y + {a_2}{y^2} + ....$$

$\Rightarrow n - m = 10$ ..... (i)

$\Rightarrow {}^m{C_2} + {}^n{C_2} - mn = 10$...... (ii)

${{m(m - 1)} \over 2} + {{n(n - 1)} \over 2} - mn = 10$

$\Rightarrow {{{m^2} - m} \over 2} + {{(10 + m)(9 + m)} \over 2} - m(10 + m) = 10$

$\Rightarrow {m^2} - m + {m^2} + 19m + 90 - 2({m^2} + 10m) = 20$

$\Rightarrow 18m + 90 - 20m = 20$

$\Rightarrow 2m = 70$

$\Rightarrow m = 35$ & $n = 45$

$m + n = 80$"

For the natural numbers m, n, if ${(1 - y)^m}{(1 + y)^n} = 1 + {a_1}y + {a_2}{y^2} + .... + {a_{m + n}}{y^{m + n}}$ and ${a_1} = {a_2} = 10$, then the value of (m + n) is equal to :

Solution

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For the natural numbers m, n, if ${(1 - y)^m}{(1 + y)^n} = 1 + {a_1}y + {a_2}{y^2} + .... + {a_{m + n}}{y^{m + n}}$ and ${a_1} = {a_2} = 10$, then the value of (m + n) is equal to :

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