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

The given problem involves a sequence and the summation of its terms with a specific formula. We are required to determine how many of the terms \(P_1, P_2, \ldots, P_m\), representing the first \(m\) prime numbers, constitute the given product. Let's solve the problem step-by-step:

1. We start with the given formula for the sum of sequence terms: \(a_1 + a_2 + \ldots + a_n = \frac{n^2 + 3n}{(n+1)(n+2)}\).
2. To find each term of the sequence \(a_n\), we consider the difference between consecutive sums:
   • a_n = (a_1 + a_2 + \ldots + a_n) - (a_1 + a_2 + \ldots + a_{n-1})
   • Using the given formula, we have: \(a_n = \frac{n^2 + 3n}{(n+1)(n+2)} - \frac{(n-1)^2 + 3(n-1)}{n(n+1)}\)
3. Simplifying the expression for \(a_n\):
   • = \frac{(n^2 + 3n) \cdot n(n+1) - ((n-1)^2 + 3(n-1)) \cdot (n+1)(n+2)}{n(n+1)(n+2)}\)
   • This simplifies to give \(a_n = \frac{2}{n+2}\).
4. To solve the problem, we need to calculate: \(28 \sum_{k=1}^{10} \frac{1}{a_k}\). Substituting for \(a_k\), we have: \(a_k = \frac{2}{k+2}\), therefore: \(\frac{1}{a_k} = \frac{k+2}{2}\).
5. Calculate the summation over the given range:
   • 28 \sum_{k=1}^{10} \frac{k+2}{2} = 14 \sum_{k=1}^{10} (k+2)
   • = 14(3 + 4 + \ldots + 12)
   • The sequence from 3 to 12 is an arithmetic sequence with 10 terms. The sum of an arithmetic sequence is calculated as: \(\frac{n}{2} \cdot (\text{first term} + \text{last term})\), which gives: = \frac{10}{2} \cdot (3 + 12) = 5 \cdot 15 = 75.
   • Thus, the sum becomes: 14 \cdot 75 = 1050.
6. Finally, we factor 1050 into its prime factors:
   • 1050 = 2 \cdot 3 \cdot 5^2 \cdot 7.
   • Using each distinct prime: \(2, 3, 5, 7\)
   • It uses the first 6 prime numbers \(2, 3, 5, 7, 11, 13\).

Therefore, the number of distinct prime factors, which is the value of \(m\), is 6. Hence, the answer is 6.