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JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Binomial Theorem

Let

a = 1 + \frac{^{2} C _{2}}{3 !} + \frac{^{3} C _{2}}{4 !} + \frac{^{4} C _{2}}{5 !} + \dots

,

b = 1 + \frac{^{1} C _{0} + ^{1} C _{1}}{1 !} + \frac{^{2} C _{0} + ^{2} C _{1} + ^{2} C _{2}}{2 !} + \frac{^{3} C _{0} + ^{3} C _{1} + ^{3} C _{2} + ^{3} C _{3}}{3 !} + \dots

. Then

\frac{2 b}{a ^{2}}

is equal to ____.

Answer: 8

Solution

Recognizing the series:

a = 1 + k \geq 2 \sum \frac{( 2 k )}{( k + 1 )!} = \frac{e}{2}

, and

b = 1 + k \geq 1 \sum \frac{\sum _{j = 0}^{k} ( j k )}{k !} = 1 + k \geq 1 \sum \frac{2 ^{k}}{k !} = e^{2}

.

\frac{2 b}{a ^{2}} = \frac{2 e ^{2}}{( e /2 ) ^{2}} = \frac{2 e ^{2}}{e ^{2} /4} = 8.

Answer: 8.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Integer answer

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