Let < a n > be a sequence such that a 1 + a 2 + . . . + a n = n 2 + 3 n n + 1 n + 2 . If 28 ∑ k = 1 10 1 a k = p 1 p 2 p 3 . . . p m , where p 1 , p 2 , . . . p m are the first m prime numbers, then m is equal to

Step-by-step solution

Given, S n = ∑ i = 1 n a i = n 2 + 3 n n + 1 n + 2 Now we know that, a n = S n - S n - 1 ⇒ a n = n 2 + 3 n n + 1 n + 2 - n - 1 2 + 3 n - 1 n n + 1 ⇒ a n = 4 n n + 1 n + 2 Now solving, ∑ k = 1 10 1 a k = 1 4 ∑ k = 1 10 k k + 1 k + 2 ⇒ ∑ k = 1 10 1 a k = 1 16 ∑ k = 1 10 k k + 1 k + 2 k + 3 - k - 1 k k + 1 k + 2 ⇒ ∑ k = 1 10 1 a k = 1 16 1 · 2 · 3 · 4 . - 0 + 2 · 3 · 4 · 5 - 1 · 2 · 3 · 4 . . . . + 10 · 11 &#1