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  • For the natural numbers , if <img alt="(1-y)^{\mathrm{m}}(1+y)^{\mathrm{m}}=1+\mathrm{a}_{1} \mathrm{y}+\mathrm{a}_{2} \mathrm{y}^{2}+\ldots+\mathrm{a}_{\mat

For the natural numbers \mathrm{m}, \mathrm{n}, if (1-y)^{\mathrm{m}}(1+y)^{\mathrm{m}}=1+\mathrm{a}_{1} \mathrm{y}+\mathrm{a}_{2} \mathrm{y}^{2}+\ldots+\mathrm{a}_{\mathrm{m+n}} \mathrm{y}^{\mathrm{m}+\mathrm{n}}  and a_{1}=a_{2}=10, then the value of (m+n) is equal to:
 
Option: 1 88
Option: 2 88
Option: 3 88
Option: 4 88
Option: 5 64
Option: 6 64
Option: 7 64
Option: 8 64
Option: 9 100
Option: 10 100
Option: 11 100
Option: 12 100
Option: 13 80
Option: 14 80
Option: 15 80
Option: 16 80

Answers (1)

best_answer

\begin{aligned} &(1-y)^{m}(1+y)^{n} \\ \end{aligned}

\begin{aligned} =&\left(1-m y+{ }^{m} C_{2} y^{2}+\cdots\right)\left(1+n y+{ }^{n} C_{2} \cdot y^{2}+\cdots\right) \\ =& 1+y(n-m)+y^{2}\left({ }^{n} C_{2}+{ }^{m} C_{2}-m n\right)+\cdots \end{aligned}

Given that this equals 1+a_{1} y+a_{2} y^{2}+\cdots$. with $a_{1}=a_{2}=10.

So, by comparing coeficients of y$ and $y^{2}.
\begin{aligned} & n-m=10 \\ \end{aligned}           

and

\begin{aligned} & ^n C_{2}+{ }^{m} C_{2}-m n=10 \\ \Rightarrow & \frac{n(n-1)}{2}+\frac{m(m-1)}{2}-m n=10 \end{aligned}

\begin{aligned} &\Rightarrow n^{2}-n+m^{2}-m-2 m n=20 \\ &\Rightarrow n^{2}+m^{2}-2 m n-(m+n)=20 \\ &\Rightarrow(n-m)^{2}-(m+n)=20 \\ &\Rightarrow \quad(m+n)=10^{2}-20 \\ &\Rightarrow \quad m+n=80 \end{aligned}

Hence option (4) is correct answer

Posted by

Kuldeep Maurya

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