If tₙ = 1/4 (n+2)(n+3), n N, then which one of the following is true ? Assertion (A) : 1/t₁ + 1/t₂ + . . . + {1}{t₂003}} = 2003/3009 Reason (R) : 1/t₁ + 1/t₂ + . . . + 1/tₙ = 4n/2n+3 [AP EAPCET 23 May 2025 S1]

Correct Answer is : (A) is false, (R) is false

(A) and (R) are true and (R) is a correct explanation of (A)

(A) and (R) are true, but (R) is not the correct explanation of (A)

Correct Answer is : (A) is false, (R) is false

Test Index

Sequences and Series

Section: Mathematics

Question : 5 of 25

Marks: +1, -0

t_n = \frac{1}{4} (n+2)(n+3), n \in N

\frac{1}{t_1} + \frac{1}{t_2} + . . . + \frac{1}{t_{2003}} = \frac{2003}{3009}

\frac{1}{t_1} + \frac{1}{t_2} + . . . + \frac{1}{t_n} = \frac{4n}{2n+3}

Solution:

Given,

t_n = \frac{1}{4} (n+2)(n+3), n \in N \quad \cdot\cdot\cdot\cdot\cdot\cdot\cdot (i)

Also,

\frac{1}{t_1} + \frac{1}{t_2} + . . . + \frac{1}{t_{2003}} = \frac{2003}{3009} \quad \cdot\cdot\cdot\cdot\cdot\cdot\cdot (ii)

From Eq. (i), we get

\frac{1}{t_n} = \frac{4}{(n+2)(n+3)}

= \frac{4}{n+2} - \frac{4}{n+3}

(By partial fraction)

\Rightarrow \quad \sum\limits_{n=1}^{2003} \frac{1}{t_n} = \sum\limits_{n=1}^{2003} \left( \frac{4}{n+2} - \frac{4}{n+3} \right)

= 4 \left( \frac{1}{3} - \frac{1}{2006} \right) = 4 \left( \frac{2006-3}{3 \times 2006} \right)

= 4 \times \frac{2003}{3 \times 2006}

= \frac{8012}{6018} = \frac{4006}{3009}

So, Assertion is false.

Also, by putting

n = 2003

in Reason then we observe that it is wrong.

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