The sum to infinite term of the series 1 + 3 2 + 3 2 6 + 3 3 10 + 3 4 14 + ⋯ is

We have Multiplying both sides by 3 1 , we get Subtracting eqn. (ii) from eqn. (i), we get 3 2 S = 1 + 3 1 + 3 2 4 + 3 3 4 + 3 4 4 + ……… ∞ ⇒ 3 2 S = 3 4 + 3 2 4 + 3 3 4 + 3 4 4 + ……… ∞ ⇒ 3 2 S = 1 − 3 1 3 4 = 3 4 × 2 3 ⇒ S = 3

Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesEasy2 minPYQ_2023

MathematicsEasysingle choice

The sum to infinite term of the series $1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \frac{14}{3 ^{4}} + \dots$ is

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Answer:

Solution:

We have

Multiplying both sides by $\frac{1}{3}$ , we get

Subtracting eqn. (ii) from eqn. (i), we get
$\frac{2}{3} S = 1 + \frac{1}{3} + \frac{4}{3 ^{2}} + \frac{4}{3 ^{3}} + \frac{4}{3 ^{4}} + \dots\dots\dots \infty \Rightarrow \frac{2}{3} S = \frac{4}{3} + \frac{4}{3 ^{2}} + \frac{4}{3 ^{3}} + \frac{4}{3 ^{4}} + \dots\dots\dots \infty \Rightarrow \frac{2}{3} S = \frac{\frac{4}{3}}{1 - \frac{1}{3}} = \frac{4}{3} \times \frac{3}{2} \Rightarrow S = 3$

Stream:BITSATSubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: PYQ_2023

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