MCQ
જો $\left({ }^{40} C _{0}\right)+\left({ }^{41} C _{1}\right)+\left({ }^{42} C _{2}\right)+\ldots+\left({ }^{\infty} C _{20}\right)=\frac{ m }{ n }{ }^{60} C _{20}, m$ અને $n$ પરસ્પર અવિભાજય હોય,તો $m+n=\dots\dots\dots$
  • $102$
  • B
    $103$
  • C
    $104$
  • D
    $105$

Answer

Correct option: A.
$102$
a
${ }^{40} C _{0}+{ }^{41} C _{1}+{ }^{42} C _{2}+\ldots . .{ }^{59} C _{19}+{ }^{60} C _{20}$

$\left(\frac{1}{41}+1\right){ }^{41} C _{1}+{ }^{42} C _{2}+\ldots \ldots$

$\left[\frac{42}{41}\left(\frac{2}{42}\right)+1\right]{ }^{42} C _{2}+{ }^{43} C _{3}+\ldots .$

$\left(\frac{2}{41}+1\right)^{42} C _{2}+{ }^{43} C _{3}+\ldots . .$

$\left(\frac{43}{41} \times \frac{3}{43}+1\right){ }^{43} C _{3}+{ }^{44} C _{4}+\ldots \ldots .$

$\frac{3+41}{41}{ }^{43} C _{3}+\ldots \ldots .$

Similarly :

$\frac{20+41}{41}$

$\Rightarrow m =61 ; n =41$

$m + n =102$

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