Series/Sigma help. *question below* will give medal

Question

itiaax · Answer

Can someone help me with 1. (a)(ii) of this question, please? I have already done part (i) and but I don't know how to go about part (ii).

freckles · Answer

that looks like an induction problem

freckles · Answer

Show the base case.

freckles · Answer

Then assume the statement is true for n=k
and show it is true for n=k+1

freckles · Answer

that works to

aum · Answer

$$
\sum_{r=1}^{n}\frac{2}{4r^2-1} = \sum_{r=1}^{n}\frac{1}{2r-1} - 
\sum_{r=1}^{n}\frac{1}{2r+1}= \
\frac 11 - \frac 13 + \frac 13 - \frac 15 + \frac 15 - \frac 17~ ... ~- \frac{1}{2n+1} = \
1 -  \frac{1}{2n+1} = \frac{2n+1-1}{2n+1} = \frac{2n}{2n+1}
$$

itiaax · Answer

Thank you! Just allow me to process everything

aum · Answer

You are welcome.