JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1

\left\{ {{a_n}} \right\}\;and\;\left\{ {{b_n}} \right\}

be two sequences given by

{a_n} = {\left( x \right)^{\frac{1}{{{2^n}}}}} + {\left( y \right)^{\frac{1}{{{2^n}}}}}

and

{b_n} = {\left( x \right)^{\frac{1}{{{2^n}}}}} - {\left( y \right)^{\frac{1}{{{2^n}}}}}

for all

n \in N,

then

{a_1}{a_2}{a_3} \ldots {a_n}

is equal to

View answer and explanation

Correct option: C

\frac{{x - y}}{{{b_n}}}

Option C (

\frac{{x - y}}{{{b_n}}}

) is correct because the product telescopes, simplifying to

\frac{x-y}{b_n}

using the difference of squares repeatedly.

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