Answer:
24/49
Step-by-step explanation:
Let's add the terms and see if there's a pattern
[tex]\dfrac{1}{1\times 3}+\dfrac{1}{3\times 5}=\dfrac{5+1}{1\times 3\times 5}=\dfrac{2}{5}\quad\text{sum of 2 terms}\\\\\dfrac{2}{5}+\dfrac{1}{5\times 7}=\dfrac{14+1}{5\times7}=\dfrac{3}{7}\quad\text{sum of 3 terms}[/tex]
Suppose we say the sum of n terms is (n/(2n+1)), the next term in the series will be 1/((2n+1)(2n+3)) and adding that to the presumed sum gives ...
[tex]\dfrac{n}{2n+1}+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n(2n+3)+1}{(2n+1)(2n+3)}=\dfrac{2n^2+3n+1}{(2n+1)(2n+3)}\\\\=\dfrac{(2n+1)(n+1)}{(2n+1)(2n+3)}=\dfrac{n+1}{2n+3}\text{ matches }\dfrac{(n+1)}{2(n+1)+1}[/tex]
Then it appears the sum of n terms is (n/(2n+1)). So, the sum of 24 terms is ...
[tex]S_{24}=\dfrac{24}{2\times24+1}=\boxed{\dfrac{24}{49}}[/tex]
