Let S_k = 1 + 2 + 3 + + k and Q_n = (S_2)/(S_2 - 1) · (S_3)/(S_3 - 1) · (S_4)/(S_4 - 1) (S_n)/(S_n - 1) , where k, n ∈ N . _ n → ∞ Q_n =

Correct answer: D. 0 We know S_n = (n(n+1))/(2) . (S_n)/(S_n - 1) = ((n(n+1))/(2))/((n(n+1))/(2) - 1) = (n^2 + n)/(n^2 + n - 2) = (n(n+1))/((n+2)(n-1)) = ((n)/(n-1))((n+1)/(n+2)). So Q_n can be written as the product of two terms: Q_n = ( (2)/(1) × (3)/(2) × × (n)/(n-1) ) × ( (3)/(4) × (4)/(5) × × (n+1)/(n+2) ). In the first bracket, everything cancels out except (n)/(1) =n . In the second bracket, everything cancels out except (3)/(n+2). Therefore, Q_n = n × (3)/(n+2) = (3n)/(n+2). Taking the limit as n → ∞ : _ n → ∞ Q_n = _ n → ∞ (3)/(1 + (2)/(n)) =3 .

Black Advanced Problems in Mathematics

Exam: Black Advanced Problems in Mathematics
Subject: Mathematics
Chapter: Sequences and Series

Black Advanced Problems in Mathematics · Mathematics · Sequences and Series

Let S_k = 1 + 2 + 3 + + k and Q_n = (S_2)/(S_2 - 1) · (S_3)/(S_3 -...

Let

S_k = 1 + 2 + 3 + \dots + k

and

Q_n = \frac{S_2}{S_2 - 1} \cdot \frac{S_3}{S_3 - 1} \cdot \frac{S_4}{S_4 - 1} \dots \frac{S_n}{S_n - 1}

, where

k, n \in N

\lim_{n \to \infty} Q_n =

Options

A $\frac{1}{3}$
B $1$
C $3$
D $0$

Correct answer

D. $0$

Step-by-step solution

We know

S_n = \frac{n(n+1)}{2}

\frac{S_n}{S_n - 1} = \frac{\frac{n(n+1)}{2}}{\frac{n(n+1)}{2} - 1} = \frac{n^2 + n}{n^2 + n - 2} = \frac{n(n+1)}{(n+2)(n-1)} = \left(\frac{n}{n-1}\right)\left(\frac{n+1}{n+2}\right). So

Q_n

can be written as the product of two terms:

Q_n =

\left( \frac{2}{1} \times \frac{3}{2} \times \dots \times \frac{n}{n-1} \right) \times \left( \frac{3}{4} \times \frac{4}{5} \times \dots \times \frac{n+1}{n+2} \right).

In the first bracket, everything cancels out except

\frac{n}{1} =n

. In the second bracket, everything cancels out except

\frac{3}{n+2}.

Therefore,

Q_n = n \times \frac{3}{n+2} = \frac{3n}{n+2}.

Taking the limit as

n \to \infty

\lim_{n \to \infty} Q_n = \lim_{n \to \infty} \frac{3}{1 + \frac{2}{n}} =3

Practice Sequences and Series on Quantrex Academy →

More from Sequences and Series

Let α = 3+4+8+9+13+14+… upto 40 terms. If ( β)^ (α)/(1020) is a root of the equation… JEE Main 2026 The sum 1 + (1)/(2)(1^2 + 2^2) + (1)/(3)(1^2 + 2^2 + 3^2) + … upto 10 terms is equal to : JEE Main 2026 The value of 1^3 - 2^3 + 3^3 - … + 15^3 is: JEE Main 2026 The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a… JEE Main 2026 For the functions f(θ) = α ^2θ + β ^2θ , and g(θ) = α ^2θ + β ^2θ , α > β > 0 , let _ 0 <… JEE Main 2026 If the sum of the first 10 terms of the series (1)/(1 + 1^4 × 4) + (2)/(1 + 2^4 × 4) +… JEE Main 2026 Let A_1, A_2, A_3, …, A_ 39 be 39 arithmetic means between the numbers 59 and 159 . Then… JEE Main 2026 Let the sum of the first n terms of an A.P. be 3n^2 + 5n . Then the sum of squares of the… JEE Main 2026 ∑_ n=1 ^ 10 ( (528)/(n(n+1)(n+2)) ) is equal to: JEE Main 2026 Let α=(1)/(4)+(1)/(8)+(1)/(16)+…∞ and β=(1)/(3)+(1)/(9)+(1)/(27)+…∞ . Then the value of… JEE Main 2026 The first term of an A.P. of 30 non-negative terms is (10)/(3) . If the sum of this A.P.… JEE Main 2026 Let a_1, a_2, a_3, … be an A.P. and g_1 = a_1, g_2, g_3, … be an increasing G.P. If a_1 =… JEE Main 2026 The sum (1^3)/(1) + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + ⋯ up to 8… JEE Main 2026 Let α, α + 2, α ∈ ℤ , be the roots of the quadratic equation x(x+2) + (x+1)(x+3) +… JEE Main 2026 Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common… JEE Main 2026 If ∑_ k=1 ^ n a_k = 6n^3 , then ∑_ k=1 ^ 6 ( a_ k+1 - a_k 36 )^2 is equal to _______. JEE Main 2026 Full Sequences and Series list Sequences and Series JEE Maths PYQs JEE Mathematics PYQs All Black Book Advanced Problems in Mathematics PYQs