If $\left({}^{40}\mathrm{C}_{0}\right)+\left({}^{41}\mathrm{C}_{1}\right)+\left({}^{42}\mathrm{C}_{2}\right)+\ldots+\left({}^{60}\mathrm{C}_{20}\right)=\frac{\mathrm{m}}{\mathrm{n}}{}^{60}\mathrm{C}_{20}$, $m$ and $n$ are coprime, then $m+n$ is equal to ____.

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Binomial Theorem

(^{40} C_{0}) + (^{41} C_{1}) + (^{42} C_{2}) + \dots + (^{60} C_{20}) = \frac{m}{n}^{60} C_{20}

m

and

n

are coprime, then

m + n

is equal to ____.

Answer: 102

Use the hockey-stick identity repeatedly. We have:

^{40} C_{0} +^{41} C_{1} +^{42} C_{2} + \dots +^{60} C_{20} =^{61} C_{20}

By the identity

^{n} C_{r} = \frac{n}{r} \cdot^{n - 1} C_{r - 1}

, we can write:

^{61} C_{20} = \frac{61}{41} \cdot^{60} C_{20}

Thus

\frac{m}{n} = \frac{61}{41}

. Since

g cd (61, 41) = 1

, we have

m = 61

and

n = 41

.

Therefore,

m + n = 61 + 41 = 102

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Integer answer