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

Let

(x + 10)^{50} + (x - 10)^{50} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{50} x^{50}

, for

x \in R

; then

\frac{a _{2}}{a _{0}}

is equal to:

(A)12.50
(B)12.00
(C)12.25✓ Correct
(D)12.75

Answer: (C)

Solution

Adding

(x + 10)^{50} + (x - 10)^{50}

keeps only the even powers of

x

, so

a_{0} = 2 \cdot 1 0^{50}

and

a_{2} = 2 (2 50) 1 0^{48}

. Hence

\frac{a _{2}}{a _{0}} = \frac{2 ( 2 50 ) 1 0 ^{48}}{2 \cdot 1 0 ^{50}} = \frac{( 2 50 )}{100} = \frac{50 \cdot 49}{2 \cdot 100} = \frac{2450}{200} = 12.25

.

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

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