Question

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

Answer 1

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).

Answer 2

that looks like an induction problem

Answer 3

Show the base case.

Answer 4

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

Answer 5

that works to

Answer 6

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

Answer 7

Thank you! Just allow me to process everything

Answer 8

You are welcome.