Let S k = k 1 + 2 + ⋯ + k and ∑ j = 1 n S j 2 = A n ( B n 2 + C n + D ) , where A , B , C , D ∈ N and A is least possible. Then:

S k = ( k + 1 ) /2 , so ∑ S j 2 = 4 1 [ 6 ( n + 1 ) ( n + 2 ) ( 2 n + 3 ) − 1 ] = 24 n ( 2 n 2 + 9 n + 13 ) . Thus ( A , B , C , D ) = ( 24 , 2 , 9 , 13 ) and ( A + B ) / D = 2 . The other statements are false.

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Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

MathematicsMediumsingle choice

Let $S_{k} = \frac{1 + 2 + \dots + k}{k}$ and $\sum_{j = 1}^{n} S_{j}^{2} = \frac{n}{A} (B n^{2} + C n + D)$ , where $A, B, C, D \in N$ and $A$ is least possible. Then:

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Solution:

$S_{k} = (k + 1) /2$ , so $\sum S_{j}^{2} = \frac{1}{4} [\frac{( n + 1 ) ( n + 2 ) ( 2 n + 3 )}{6} - 1] = \frac{n}{24} (2 n^{2} + 9 n + 13)$ . Thus $(A, B, C, D) = (24, 2, 9, 13)$ and $(A + B) / D = 2$ . The other statements are false.

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

⏱ 2mℹ️ Source: QB

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