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JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Binomial Theorem

Let

n \in N

and

[x]

denote the greatest integer less than or equal to

x

. If the sum of (

n + 1

) terms

^{n} C_{0}, 3.^{n} C_{1}, 5.^{n} C_{2}, 7.^{n} C_{3}, \dots .

. is equal to

2^{100} .101

, then

2 [\frac{n - 1}{2}]

is equal to

Answer: 98

Solution

The sum is

S = \sum_{r = 0}^{n} (2 r + 1) (r n)

.

We can split this:

S = \sum_{r = 0}^{n} 2 r (r n) + \sum_{r = 0}^{n} (r n)

Using the identity

r (r n) = n (r - 1 n - 1)

:

\sum_{r = 0}^{n} 2 r (r n) = 2 n \sum_{r = 1}^{n} (r - 1 n - 1) = 2 n \cdot 2^{n - 1} = n \cdot 2^{n}

And

\sum_{r = 0}^{n} (r n) = 2^{n}

Therefore:

S = n \cdot 2^{n} + 2^{n} = (n + 1) \cdot 2^{n}

Given

S = 2^{100} \cdot 101

:

(n + 1) \cdot 2^{n} = 2^{100} \cdot 101

This gives

n + 1 = 101

and

2^{n} = 2^{100}

, so

n = 100

.

Therefore:

2 [\frac{n - 1}{2}] = 2 [\frac{99}{2}] = 2 [49.5] = 2 \cdot 49 = 98

.

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Integer answer

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