Let { a n } satisfy a 1 + ⋯ + a n = ( n + 1 ) ( n + 2 ) n 2 + 3 n . If 28 ∑ k = 1 10 a k 1 = p 1 p 2 ⋯ p m , where the p i are the first m primes, then m equals:

Subtracting consecutive partial sums gives a n = 4/ [ n ( n + 1 ) ( n + 2 )] . Therefore 28 ∑ k = 1 10 1/ a k = 7 ∑ k = 1 10 k ( k + 1 ) ( k + 2 ) = 4 7 ( 10 ⋅ 11 ⋅ 12 ⋅ 13 ) = 30030 = 2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 . Hence m = 6 .

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MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

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Let ${a_{n}}$ satisfy $a_{1} + \dots + a_{n} = \frac{n ^{2} + 3 n}{( n + 1 ) ( n + 2 )}$ . If $28 \sum_{k = 1}^{10} \frac{1}{a _{k}} = p_{1} p_{2} \dots p_{m}$ , where the $p_{i}$ are the first $m$ primes, then $m$ equals:

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Subtracting consecutive partial sums gives $a_{n} = 4/ [n (n + 1) (n + 2)]$ . Therefore $28 \sum_{k = 1}^{10} 1/ a_{k} =$ $7 \sum_{k = 1}^{10} k (k + 1) (k + 2) = \frac{7}{4} (10 \cdot 11 \cdot 12 \cdot 13) = 30030 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ . Hence $m = 6$ .

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

⏱ 2mℹ️ Source: QB

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