Let be a sequence such that a1+a2+....+ an=\(\frac{n^2+3n}{ (n+1 ) (n+2)}\). If \(28∑^{10}_{k=1}\frac{1}{a_k} =\) P1P2P3 . . . Pm , where p1, p2, …..pm are the first m prime numbers, then m is equal to
- 5
- 6
- 7
- 8
The Correct Option is B
Solution and Explanation
Given: \[ a_n = S_n - S_{n-1} = n^2 + 3n. \]
Simplify the expression: Using the formula for \(a_n\): \[ a_n = \frac{(n+1)(n+2) - (n-1)(n+2)}{n(n+1)}. \]
Now, consider: Summing over \(k\) from 1 to 10: \[ \sum_{k=1}^{10} a_k = \sum_{k=1}^{10} \frac{k(k+1)(k+2)}{4}. \]
Expand and calculate: Substitute \(k(k+1)(k+2)\) and simplify further using telescoping: \[ \sum_{k=1}^{10} \frac{(k(k+1)(k+2)(k+3) - (k-1)k(k+1)(k+2))}{4}. \]
After calculation: The final result is: \[ \sum_{k=1}^{10} a_k = \frac{7}{4} \cdot 10 \cdot 11 \cdot 12 \cdot 13 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13. \]
Final Answer: The value of \(m\) is 6.
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