Sum 5 2 ⋅ 2 ⋅ 1 9 + 5 3 ⋅ 3 ⋅ 2 13 + 5 4 ⋅ 4 ⋅ 3 17 + ⋯ to infinity.

The r th term ( r ≥ 2 ) is ( 4 r + 1 ) / ( 5 r r ( r − 1 ) ) = 1/ [ 5 r − 1 ( r − 1 ) ] − 1/ ( 5 r r ) . Summation telescopes to 1/5.

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MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

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Sum $\frac{9}{5 ^{2} \cdot 2 \cdot 1} + \frac{13}{5 ^{3} \cdot 3 \cdot 2} + \frac{17}{5 ^{4} \cdot 4 \cdot 3} + \dots$ to infinity.

Answer:

Solution:

The $r$ th term $(r \geq 2)$ is $(4 r + 1) / (5^{r} r (r - 1)) = 1/ [5^{r - 1} (r - 1)] - 1/ (5^{r} r)$ . Summation telescopes to 1/5.

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

⏱ 2mℹ️ Source: QB

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