Question

$\frac{1}{1.5}+\frac{1}{5.9}+\dots .+\frac{1}{(4n-3)(4n+1)}=$

• A. $\frac{n}{4n-3}$
• B. $\frac{n}{4n+3}$
• C. $\frac{n}{4n+1}$
• D. $\frac{n}{4n+5}$

Answer 1

Answer: (C)

$\frac{n}{4n+1}$

Consider the $i^{th}$ term: $t_i=\frac{1}{(4i-3)(4i+1)}$.
We re-write it as :
$t_i=\frac{1}{4}\times \frac{(4i+1)-(4i-3)}{(4i-3)(4i+1)}=\frac{1}{4}\times (\frac{1}{4i-3}-\frac{1}{4i+1})$
Similarly, $t_{i+1}=\frac{1}{4}\times (\frac{1}{4i+1}-\frac{1}{4i+5})$
Observe that first time in $t_i$ matches with the second term of $t_{i-1}$, while second term of $t_i$ matches with the first term of $t_{i+1}$. This is an instance of telescoping summing. Only the first term of $t_1$ and the last term of $t_n$ will remain.
Answer = $\frac{1}{4}\times (1-\frac{1}{4n+1})=\frac{n}{4n+1}$