IIT JAM 2021 Question 45: Difference and differential equations

Consider $x_n = \left(\dfrac{n+1}{n}\right)x_{n-1}$ for $n = 1, 2, 3, \ldots$ with $x_0 = 2$ . Then $x_{100} - x_{50}$ equals _______ (in integer).

Answer and worked solution

Answer: 100

Iterating, $x_n = x_0 \prod_{k=1}^n \dfrac{k+1}{k}$ , and the product telescopes to $(n+1)$ , so $x_n = 2(n+1)$ .

Then $x_{100} - x_{50} = 2(101) - 2(51) = 202 - 102 = 100$ . The answer is $100$ .

Plain-text version

Consider xₙ = ((n+1)/n)xₙ₋₁ for n = 1, 2, 3, … with x₀ = 2. Then x₁₀₀ - x₅₀ equals _ (in integer).

Also written as: Consider x_n = ((n+1)/n)x_(n-1) for n = 1, 2, 3,... with x_0 = 2. Then x_(100) - x_(50) equals _ (in integer).

Difference and differential equations has come up 20 times in 5 exams since 2014.

This question is on Difference equations, part of Difference and differential equations in Mathematics.

Solving equations that link a variable to its past values.

TestStop has 5 past-paper questions on this subtopic; the nearest in year are below.

Question from the IIT JAM 2021 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.