MCQ
If $\left({ }^{40} C _{0}\right)+\left({ }^{41} C _{1}\right)+\left({ }^{42} C _{2}\right)+\ldots+\left({ }^{\infty} C _{20}\right)=\frac{ m }{ n }{ }^{60} C _{20}, m$ and $n$ are coprime, then $m+n$ is equal to
- ✓$102$
- B$103$
- C$104$
- D$105$
$\left(\frac{1}{41}+1\right){ }^{41} C _{1}+{ }^{42} C _{2}+\ldots \ldots$
$\left[\frac{42}{41}\left(\frac{2}{42}\right)+1\right]{ }^{42} C _{2}+{ }^{43} C _{3}+\ldots .$
$\left(\frac{2}{41}+1\right)^{42} C _{2}+{ }^{43} C _{3}+\ldots . .$
$\left(\frac{43}{41} \times \frac{3}{43}+1\right){ }^{43} C _{3}+{ }^{44} C _{4}+\ldots \ldots .$
$\frac{3+41}{41}{ }^{43} C _{3}+\ldots \ldots .$
Similarly :
$\frac{20+41}{41}$
$\Rightarrow m =61 ; n =41$
$m + n =102$
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$(A)$ $\int^{\pi / 4} x f(x) d x=\frac{1}{12}$
$(B)$ $\int_0^{\pi / 4} f(x) d x=0$
$(C)$ $\int_0^{\pi / 4} x f(x) d x=\frac{1}{6}$
$(D)$ $\int_0^{\pi / 4} f(x) d x=1$