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JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Binomial Theorem

Let

S_{n} = 1 + q + q^{2} + \dots + q^{n}

and

T_{n} = 1 + (\frac{q + 1}{2}) + (\frac{q + 1}{2})^{2} + \dots + (\frac{q + 1}{2})^{n}

where

q

is a real number and

q \neq = 1

. If

^{101} C_{1} +^{101} C_{2} \cdot S_{1} + \dots +^{101} C_{101} \cdot S_{100} = α T_{100}

then

α

is equal to:

(A) $2^{99}$
(B)202
(C)200
(D) $2^{100}$ ✓ Correct

Answer: (D)

Solution

With

S_{r - 1} = \frac{q ^{r} - 1}{q - 1}

,

\sum_{r = 1}^{101}^{101} C_{r} S_{r - 1} = \frac{1}{q - 1} (\sum_{r = 1}^{101}^{101} C_{r} q^{r} - \sum_{r = 1}^{101}^{101} C_{r}) = \frac{1}{q - 1} ((1 + q)^{101} - 1 - (2^{101} - 1)) = \frac{( 1 + q ) ^{101} - 2 ^{101}}{q - 1}

. Also

T_{100} = \frac{( \frac{q + 1}{2} ) ^{101} - 1}{\frac{q + 1}{2} - 1} = \frac{( q + 1 ) ^{101} - 2 ^{101}}{2 ^{100} ( q - 1 )}

. Comparing, the sum equals

α T_{100}

when

α = 2^{100}

.

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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