Answer:
The statement is true
Step-by-step explanation:
We will prove by mathematical induction that, for every natural n,
[tex]\sum^{n}_{i=1}\frac{1}{(2i-1)(2i+1)} =\frac{n}{2n+1}[/tex]
We will prove our base case, when n=1, to be true.
base case:
[tex]\sum^{1}_{i=1}\frac{1}{(2-1)(2+1)} =\frac{1}{3}=\frac{n}{2n+1}[/tex]
Inductive hypothesis:
[tex]\sum^{n}_{i=1}\frac{1}{(2i-1)(2i+1)} =\frac{n}{2n+1}[/tex]
Now, we will assume the induction hypothesis and then uses this assumption, involving n, to prove the statement for n + 1.
Inductive step:
[tex]\sum^{n+1}_{i=1}\frac{1}{(2i-1)(2i+1)} =\sum^{n}_{i=1}\frac{1}{(2i-1)(2i+1)}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}=\frac{n(2n+3)+1}{(2n+1)(2n+3)}=\frac{2n^2+3n+1}{(2n+1)(2n+3)}=\frac{(2n+1)(n+1)}{(2n+1)(2n+3)}=\frac{n+1}{2n+3}=\frac{n+1}{2(n+1)+1}[/tex]
With this we have proved our statement to be true for n+1.
In conlusion, for every natural [tex]n[/tex].
[tex]\sum^{n}_{i=1}\frac{1}{(2i-1)(2i+1)} =\frac{n}{2n+1}[/tex]
