TS EAMCET · Maths · Functions
If \(t_n=\frac{1}{4}(n+2)(n+3)\) for \(n=1,2,3 \ldots\), then \(\frac{1}{t_1}+\frac{1}{t_2}+\ldots+\frac{1}{t_{2003}}\) is equal to
- A \(\frac{4006}{3006}\)
- B \(\frac{4003}{3007}\)
- C \(\frac{4006}{3008}\)
- D \(\frac{4006}{3009}\)
Answer & Solution
Correct Answer
(D) \(\frac{4006}{3009}\)
Step-by-step Solution
Detailed explanation
We have, \(t_n=\frac{1}{4}(n+2)(n+3)\) \(\frac{1}{t_n}=\frac{4}{(n+2)(n+3)}\) \(=4\left[\frac{1}{n+2}-\frac{1}{n+3}\right]\) Put \(n=1,2,3, \ldots, 2003\) \(\frac{1}{t_1}=4\left(\frac{1}{3}-\frac{1}{4}\right)\)…
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