Question 11
A sequence $$\{a_n\}$$, $$n \geq 1$$ with $$a_1 = \frac{1}{2}$$ and $$a_n = \frac{a_{n-1}}{2na_{n-1} + 1}$$ is given. Then the value of $$a_1 + a_2 + a_3 + \ldots + a_{2024}$$ is equal to
Solution
Taking reciprocals, $$\frac{1}{a_n} = \frac{1}{a_{n-1}} + 2n$$ with $$\frac{1}{a_1} = 2$$, which gives $$\frac{1}{a_n} = n(n+1)$$. So $$a_n = \frac{1}{n} - \frac{1}{n+1}$$ and the sum telescopes to $$1 - \frac{1}{2025} = \frac{2024}{2025}$$.
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