JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(\left\langle a _{ n }\right\rangle\) be a sequence such that \(a_1+a_2+\ldots+a_n=\frac{n^2+3 n}{(n+1)(n+2)}\). If \(28 \sum \limits_{ k =1}^{10} \frac{1}{ a _{ k }}= p _1 p _2 p _3 \ldots p _{ m }\), where \(p _1, p _2, \ldots . pm\) are the first \(m\) prime numbers, then \(m\) is equal to
- A \(7\)
- B \(6\)
- C \(5\)
- D \(8\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
\(a_n=S_n-S_{n-1}=\frac{ n ^2+3 n }{( n +1)( n +2)}-\frac{( n -1)( n +2)}{ n ( n +1)}\) \(\Rightarrow a _{ n }=\frac{4}{ n ( n +1)( n +2)}\) \(\Rightarrow 28 \sum \limits_{ k =1}^{10} \frac{1}{ a _{ k }}=28 \sum \limits_{ k =1}^{10} \frac{ k ( k +1)( k +2)}{4}\)…
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