If \(1+\left(2+{ }^{49} \mathrm{C}_{1}+{ }^{49} \mathrm{C}_{2}+\ldots . .+{ }^{49} \mathrm{C}_{49}\right)\left({ }^{50} \mathrm{C}_{2}+{ }^{50} \mathrm{C}_{4}+\ldots . .+{ }^{50} \mathrm{C}_{50}\right)\) is equal to \(2^{\mathrm{n}} . \mathrm{m}\), where m is odd, then n +m is equal to

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Binomial Theorem

1 + (2 +^{49} C_{1} +^{49} C_{2} + \dots .. +^{49} C_{49}) (^{50} C_{2} +^{50} C_{4} + \dots .. +^{50} C_{50})

is equal to

2^{n} . m

, where m is odd, then n +m is equal to

Answer: 99

We have

\sum_{k = 0}^{49} (k 49) = 2^{49}

, so

2 + \sum_{k = 1}^{49} (k 49) = 2 + 2^{49} - 1 = 1 + 2^{49}

. Also,

\sum_{k = 0}^{50} (k 50) = 2^{50}

, and by symmetry, the sum of even-indexed binomial coefficients equals the sum of odd-indexed ones, so

\sum_{k even} (k 50) = 2^{49}

. But we need

(2 50) + (4 50) + \dots + (50 50) = 2^{49} - 1

(excluding

(0 50) = 1

). Therefore, the expression becomes

1 + (1 + 2^{49}) (2^{49} - 1) = 1 + 2^{49} \cdot 2^{49} - 1 = 2^{98}

. Thus

n = 98

and

m = 1

, giving

n + m = 99

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer