Let a 1 = 2 and a n − a n − 1 = n 2 for n ≥ 2 . If a k = 205 , find k .

Summing from 2 to k gives a k − a 1 = 2 2 + 3 2 + ⋯ + k 2 . Thus 1 2 + 2 2 + ⋯ + k 2 = 204 , so k ( k + 1 ) ( 2 k + 1 ) /6 = 204 and k = 8 .

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MathematicsSequence & SeriesSpecial sequences/seriesEasy2 minQB

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Let $a_{1} = 2$ and $a_{n} - a_{n - 1} = n^{2}$ for $n \geq 2$ . If $a_{k} = 205$ , find $k$ .

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Solution:

Summing from 2 to $k$ gives $a_{k} - a_{1} = 2^{2} + 3^{2} + \dots + k^{2}$ . Thus $1^{2} + 2^{2} + \dots + k^{2} = 204$ , so $k (k + 1) (2 k + 1) /6 = 204$ and $k = 8$ .

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

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