AP EAMCET · Maths · Sequences and Series
If \(\mathrm{t}_{\mathrm{n}}=\frac{1}{4}(\mathrm{n}+2)(\mathrm{n}+3), \mathrm{n} \in \mathrm{N}\), then which one of the following is true?
Assertion (A) : \(\frac{1}{t_1}+\frac{1}{t_2}+\ldots+\frac{1}{t_{2003}}=\frac{2003}{3009}\)
Reason (R) : \(\frac{1}{\mathrm{t}_1}+\frac{1}{\mathrm{t}_2}+\ldots+\frac{1}{\mathrm{t}_{\mathrm{n}}}=\frac{4 \mathrm{n}}{(2 \mathrm{n}+3)}\)
- A (A) and (R) are true and (R) is a correct explanation of (A)
- B (A) and (R) are true, but (R) is not the correct explanation of (A)
- C (A) is true, (R) is false
- D (A) is false, (R) is false
Answer & Solution
Correct Answer
(D) (A) is false, (R) is false
Step-by-step Solution
Detailed explanation
\(\frac{1}{t_n} = \frac{4}{(n+2)(n+3)} = 4 \left( \frac{1}{n+2} - \frac{1}{n+3} \right)\) \(\sum_{k=1}^{n} \frac{1}{t_k} = 4 \sum_{k=1}^{n} \left( \frac{1}{k+2} - \frac{1}{k+3} \right) = 4 \left( \frac{1}{3} - \frac{1}{n+3} \right) = \frac{4n}{3(n+3)}\) Reason (R):…
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