The sum (1)/(1·3)+(1)/(3·5)+(1)/(5·7)+… up to 20 terms equals

The general term is $\dfrac{1}{(2k-1)(2k+1)}$. Use partial fractions: $\dfrac{1}{(2k-1)(2k+1)}=\dfrac12\left(\dfrac{1}{2k-1}-\dfrac{1}{2k+1}\right)$. Summing from $k=1$ to $n$, the middle terms cancel (telescoping) and $S_n=\dfrac12\left(1-\dfrac{1}{2n+1}\right)=\dfrac{n}{2n+1}$. For $n=20$ this gives $S_{20}=\dfrac{20}{41}$. The tempting wrong answers come from using $n-1=19$ terms ($\tfrac{19}{39}$), the wrong number of terms ($\tfrac{10}{21}$ is the sum of $10$ terms) or forgetting the factor $\tfrac12$ ($\tfrac{40}{41}$). Check with $n=1$: $\tfrac13$, matching the first term.