Find 1/9 + 1/18 + 1/30 + 1/45 + 1/63 + ⋯ to infinity.

The k th denominator is 3 ( k + 1 ) ( k + 2 ) /2 , so the sum is 3 2 ∑ k = 1 ∞ [ 1/ ( k + 1 ) − 1/ ( k + 2 )] = 1/3 .

Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

MathematicsMediumsingle choice

Find $1/9 + 1/18 + 1/30 + 1/45 + 1/63 + \dots$ to infinity.

Answer:

Solution:

The $k$ th denominator is $3 (k + 1) (k + 2) /2$ , so the sum is $\frac{2}{3} \sum_{k = 1}^{\infty} [1/ (k + 1) - 1/ (k + 2)] = 1/3$ .

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

⏱ 2mℹ️ Source: QB

Loading discussions...