Find the sum \(\frac{1}{4}+\frac{3}{16}+\frac{5}{64}+\frac{7}{256}+...+\frac{(2n-1)}{4^n}+...\)
As follows:
a(n) =( 2^(-2 n) (2 n - 1))
Let n = 100
∑[( 2^(-2 n) (2 n - 1)), n, 1, 100] = 5/9
Nice, Alan !!!
Thanks Chris!
+816 
