Prove that ∑ from k =1 to n of 1/(4k^2 - 1) = n/(2n+1) for all positive integers n.
In words: Prove that the sum (from k =1 to n) of one over (4 times k squared minus one) is equal to n over two times n + 1
I am beyond beyond beyond lost.
Prove that ∑ from k =1 to n of 1/(4k^2 - 1) = n/(2n+1) for all positive integers n.
In words: Prove that the sum (from k =1 to n) of one over (4 times k squared minus one) is equal to n over two times n + 1
I am beyond beyond beyond lost.
Samething
$\frac{1}{2}\left[ {\left( {\frac{1}{1} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{5}} \right) + \left( {\frac{1}{5} - \frac{1}{7}} \right) \cdots + \left( {\frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right)} \right] = \frac{1}{2}\left[ {1 - \frac{1}{{2n + 1}}} \right]$