Let Tr = (2r-1)(2r+1)(2r+3)(2r+5)/64, then

Question

Let \(T_r=\frac{(2 r-1)(2 r+1)(2 r+3)(2 r+5)}{64},\) then \(\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r}\) is equal to

(1) \(\frac{22}{45}\)

(2) \(\frac{32}{35}\)

(3) \(\frac{27}{45}\)

(4) \(\frac{32}{45}\)

Answer 1

Correct option is (4) \(\frac{32}{45}\)    

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

\(\Rightarrow \frac{1}{T_r}=\frac{64}{16\left(r-\frac{1}{2}\right)\left(r+\frac{1}{2}\right)\left(r+\frac{3}{2}\right)\left(r+\frac{5}{2}\right)}\)  

\(\Rightarrow \frac{1}{T_r}=\frac{\frac{4}{3}\left[\left(r+\frac{5}{2}\right)-\left(r-\frac{1}{2}\right)\right]}{\left(r-\frac{1}{2}\right)\left(r+\frac{1}{2}\right)\left(r+\frac{3}{2}\right)\left(r+\frac{5}{2}\right)}\)   

\(\Rightarrow \frac{1}{T_r}=\frac{4}{3}\left[\frac{1}{\left(r-\frac{1}{2}\right)\left(r+\frac{1}{2}\right)\left(r-\frac{3}{2}\right)}-\right. \left.\frac{1}{\left(r+\frac{1}{2}\right)\left(r+\frac{3}{2}\right)\left(r+\frac{5}{2}\right)}\right]\)  

\(\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r}=\frac{4}{3}\left[\frac{1}{\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2}}-\frac{1}{\frac{3}{2} \cdot \frac{5}{2} \cdot \frac{7}{2}}\right] \frac{1}{\frac{3}{2} \cdot \frac{5}{2}-\frac{7}{2}}-\frac{1}{\frac{5}{2} \cdot \frac{7}{2} \cdot \frac{9}{2}}\)  

\(=\frac{4}{3}\left[\frac{8}{15}\right]=\frac{32}{45}\)