2 + 4 + 7 + 11 + 16 + … to n terms =

Differences of the terms are 2 , 3 , 4 , … , so T n = 2 + ∑ k = 2 n k = 2 + 2 n ( n + 1 ) − 1 = 2 n 2 + n + 2 . S n = ∑ T n = 2 1 [ 6 n ( n + 1 ) ( 2 n + 1 ) + 2 n ( n + 1 ) + 2 n ] = 12 n [( n + 1 ) ( 2 n + 1 ) + 3 ( n + 1 ) + 12 ] = 12 n ( 2 n 2 + 6 n + 16 ) = 6 n ( n 2 + 3 n + 8 ) .

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$2 + 4 + 7 + 11 + 16 + \dots$ to $n$ terms =

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Solution:

Differences of the terms are $2, 3, 4, \dots$ , so $T_{n} = 2 + \sum_{k = 2}^{n} k = 2 + \frac{n ( n + 1 )}{2} - 1 = \frac{n ^{2} + n + 2}{2}$ . $S_{n} = \sum T_{n} = \frac{1}{2} [\frac{n ( n + 1 ) ( 2 n + 1 )}{6} + \frac{n ( n + 1 )}{2} + 2 n] = \frac{n}{12} [(n + 1) (2 n + 1) + 3 (n + 1) + 12] = \frac{n}{12} (2 n^{2} + 6 n + 16) = \frac{n}{6} (n^{2} + 3 n + 8) .$

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

⏱ 2mℹ️ Source: QB

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