If C0 40 + C1 41 + C2 42 + ⋯ + C20 60 = (m)/(n) × C20 60 where m &...

We know $Cr - 1 n + Cr n = Cr n + 1$

Also $C0 40 = C0 41$

So $C0 41 + C1 41 + C2 42 + \ldots . C20 60$

$= C1 42 + C2 42 + C3 43 . . . . C20 60$ $= C2 43 + C3 43 + \ldots C20 60$

$= C19 60 + C20 60 = C20 61$

Given $C20 61 = \dfrac{m}{n} C20 60$

$⇒ C41 61 = \dfrac{61}{41} C40 60$

$⇒ \dfrac{61}{41} C20 60 = \dfrac{m}{n} C20 60$

$⇒ \dfrac{61}{41} = \dfrac{m}{n} ⇒ m + n = 102$