Let \(\langle a_n \rangle\) be a sequence such that \(a_{1}+a_{2}+\ldots+a_{n}=\frac{n^{2}+3 n}{(n+1)(n+2)}\). If \(28 \sum_{k=1}^{10} \frac{1}{a_k}=p_{1} p_{2} p_{3} \ldots p_{m}\), where \(p_{1}, p_{2}, \ldots, p_m\) are the first \(m\) prime numbers, then \(m\) is equal to

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Sequences and Series

Let

⟨ a_{n} ⟩

be a sequence such that

a_{1} + a_{2} + \dots + a_{n} = \frac{n ^{2} + 3 n}{( n + 1 ) ( n + 2 )}

. If

28 \sum_{k = 1}^{10} \frac{1}{a _{k}} = p_{1} p_{2} p_{3} \dots p_{m}

, where

p_{1}, p_{2}, \dots, p_{m}

are the first

m

prime numbers, then

m

is equal to

Answer: (B)

With

S_{n} = \frac{n ^{2} + 3 n}{( n + 1 ) ( n + 2 )} = \frac{n ( n + 3 )}{( n + 1 ) ( n + 2 )}

, the term is

a_{n} = S_{n} - S_{n - 1}

. Using

S_{n - 1} = \frac{( n - 1 ) ( n + 2 )}{n ( n + 1 )}

and a common denominator, the numerator becomes

n^{2} (n + 3) - (n - 1) (n + 2)^{2} = 4

, so

a_{n} = \frac{4}{n ( n + 1 ) ( n + 2 )}

. Hence

\frac{1}{a _{k}} = \frac{k ( k + 1 ) ( k + 2 )}{4}

and

28 k = 1 \sum 10 \frac{1}{a _{k}} = 7 k = 1 \sum 10 k (k + 1) (k + 2)

. Since

k = 1 \sum n k (k + 1) (k + 2) = \frac{n ( n + 1 ) ( n + 2 ) ( n + 3 )}{4}

, for

n = 10

this equals

\frac{10 \cdot 11 \cdot 12 \cdot 13}{4} = 4290

, giving

7 \times 4290 = 30030

. Factoring,

30030 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13

, which is the product of the first six primes, so

m = 6

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option