Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f(n) = \left[\sqrt{n} + \frac{1}{2}\right]$, where $[.]$ denotes greatest integer function, $\forall n \in \mathbb{N}$. Then $\sum_{n=1}^{\infty} \frac{2^{f(n)} + 2^{-f(n)}}{2^n}$ is equal to______.
Step-by-Step Solution
Key Concept: Partitioning the infinite sum based on which integer interval $\sqrt{n}$ falls in reveals a telescoping structure that simplifies the computation.
Let $\sqrt{n} - \frac{1}{2} = K$ where $K$ is a positive integer satisfying $(K - 1/2)^2 \leq n < (K + 1/2)^2$, giving $K^2 - K + 1/4 \leq n \leq K^2 + K$. The sum $S = \sum_{n=1}^{\infty} \frac{2^{f(n)} + 2^{-f(n)}}{2^n}$ is reorganized by grouping terms for each value of $K$. Computing the sum yields telescoping series that converge to $S = 3$.
Correct Answer: 5