The product 2 1 4 ∙ 4 1 16 ∙ 8 1 48 ∙ 16 1 128 ∙ . . . . to ∞ is equal to:

2 1 4 + 2 16 + 3 48 + … … … . ∞ S = 2 1 4 + 1 8 + 1 16 + … . . ∞ . . . . i ∴ 1 4 + 1 8 + 1 16 + . . . ∞ is an infinite G.P with a = 1 4 , r = 1 2 S ∞ = a 1 - r = 1 2 Using equation i , S = 2

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Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesG.P.Easy2 minPYQ_2020

MathematicsEasystatement

The product $2^{\frac{1}{4}} ∙ 4^{\frac{1}{16}} ∙ 8^{\frac{1}{48}} ∙ 16^{\frac{1}{128}} ∙ . . . .$ to $\infty$ is equal to:

Options:

Answer:

Solution:

$2^{\frac{1}{4} + \frac{2}{16} + \frac{3}{48} + \dots \dots \dots . \infty}$
$S = 2^{\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots . . \infty} . . . . (i)$

$∴ \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + . . . \infty$ is an infinite G.P with $a = \frac{1}{4}, r = \frac{1}{2}$

$S_{\infty} = \frac{a}{1 - r} = \frac{1}{2}$

Using equation $(i)$ , $S = \sqrt{2}$

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:G.P.

⏱ 2mℹ️ Source: PYQ_2020

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