If r=1^n Tr = (2n-1)(2n+1)(2n+3)(2n+5)64, then n r=1^n 1Tr is equal to:

Given: r=1^n Tr = (2n-1)(2n+1)(2n+3)(2n+5)64 Find: n r=1^n 1Tr Let Sn = (2n-1)(2n+1)(2n+3)(2n+5)64 Then Tr = Sr - Sr-1 Using the expression from the solution, Tr = (2r-1)(2r+1)(2r+3)8 Therefore, 1Tr = 8(2r-1)(2r+1)(2r+3) Now use partial fractions: 8(2r-1)(2r+1)(2r+3) = 12r-1- 22r+1+ 12r+3 So, r=1^n 1Tr = r=1^n ( 12r-1- 22r+1+ 12r+3) This telescopes, and on simplifying we get r=1^n 1Tr= 8n(n+2)3(4n^2+8n+3) Taking the limit as n, n 8n(n+2)3(4n^2+8n+3)= 83 4= 23 Therefore, the correct option is C. Given: r=1^n Tr = (2n-1)(2n+1)(2n+3)(2n+5)64 Find: n r=1^n 1Tr From the extracted working, Tr = (2r-1)(2r+1)(2r+3)8 Hence 1Tr = 8(2r-1)(2r+1)(2r+3) The summand can be viewed in telescoping form, so the series collapses after cancellation. The simplified finite sum is r=1^n 1Tr= 8n(n+2)3(4n^2+8n+3) Now compare leading terms in numerator and denominator. Thus, n r=1^n 1Tr = n 8n(n+2)3(4n^2+8n+3) = 23 Therefore, the value of the limit is 23.

If r=1^n Tr = (2n-1)(2n+1)(2n+3)(2n+5)64, then n r=1^n 1Tr is equ… — Solution | JEE Mathematics

MCQMediumJEE 2025Sum of Series

If $\sum_{r=1}^n T_r = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$ , then $\lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r}$ is equal to :

A
$1$
B
$0$
C
$\frac{2}{3}$
D
$\frac{1}{3}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

\sum_{r=1}^n T_r = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}

Find:

\lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r}

Let

S_n = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}

Then

T_r = S_r - S_{r-1}

Using the expression from the solution,

T_r = \frac{(2r-1)(2r+1)(2r+3)}{8}

Therefore,

\frac{1}{T_r} = \frac{8}{(2r-1)(2r+1)(2r+3)}

Now use partial fractions:

\frac{8}{(2r-1)(2r+1)(2r+3)} = \frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}

So,

\sum_{r=1}^{n}\frac{1}{T_r} = \sum_{r=1}^{n}\left(\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}\right)

This telescopes, and on simplifying we get

\sum_{r=1}^{n}\frac{1}{T_r}=\frac{8n(n+2)}{3(4n^2+8n+3)}

Taking the limit as $n \to \infty$ ,

\lim_{n\to\infty}\frac{8n(n+2)}{3(4n^2+8n+3)}=\frac{8}{3\cdot 4}=\frac{2}{3}

Therefore, the correct option is C.

Telescoping Observation

Given:

\sum_{r=1}^n T_r = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}

Find:

\lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r}

From the extracted working,

T_r = \frac{(2r-1)(2r+1)(2r+3)}{8}

Hence

\frac{1}{T_r} = \frac{8}{(2r-1)(2r+1)(2r+3)}

The summand can be viewed in telescoping form, so the series collapses after cancellation. The simplified finite sum is

\sum_{r=1}^{n}\frac{1}{T_r}=\frac{8n(n+2)}{3(4n^2+8n+3)}

Now compare leading terms in numerator and denominator.

Thus,

\lim_{n\to\infty}\sum_{r=1}^{n}\frac{1}{T_r} = \lim_{n\to\infty}\frac{8n(n+2)}{3(4n^2+8n+3)} = \frac{2}{3}

Therefore, the value of the limit is $\frac{2}{3}$ .

Common mistakes

Computing $T_r$ incorrectly from $S_r-S_{r-1}$ . This is wrong because the entire series depends on the exact form of $T_r$ . First write $S_r$ and $S_{r-1}$ carefully, then subtract before taking the reciprocal.
Missing the partial fraction or telescoping structure in $\frac{8}{(2r-1)(2r+1)(2r+3)}$ . This is wrong because direct summation is not feasible. Rewrite the term into simpler fractions so cancellation becomes visible.
Taking the limit termwise too early. This is wrong because $\lim_{n\to\infty}\sum_{r=1}^n \frac{1}{T_r}$ must be found from the simplified finite sum, not by replacing $n$ with infinity before summing.

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