Question

Let be a sequence such that a1+a₂+....+ a_n=\(\frac{n^2+3n}{ (n+1 ) (n+2)}\). If  \(28∑^{10}_{k=1}\frac{1}{a_k} =\) P₁P₂P₃ . . . P_m ,  where p1, p2, …..p_m are the first m prime numbers, then m is equal to

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

The Correct Option is B

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.