21okay..easier one..[6]
the given limit equals
\sum_{r=1}^{\propto }{}\frac{2^{4r}-1}{2^{r(r+2)}}
which is equal to 3 ?
going to sleep..thnx for the q
5 Answers
21tell me fast if that involves definite inregrals [4]
i will try otherwise..
21
62yup subhodeep good work :)
Do you want to give a hint about how many integers there are between two consecutive perfect squares? ;)
21<n> = \left[\sqrt{n}+\frac{1}{2} \right]
<n> will be equal to a positive integer r for exactly 2r number of positive integers.And those numbers are r(r-1) + k, where k is from the set [1,2,...,2r]. So the given expression equals \sum_{r=1}^{}{\propto }\ \left([2^r + \frac{1}{2^r}][\sum_{k=1}^{k=2r}{}\frac{1}{r(r-1)+k}] \right)
=\sum_{r=1}^{}{\propto } \frac{2^{4r}-1}{2^r(r+2)} = f(r) - f(r+2)
where
f(r) = \frac{1}{2^{r(r-2)}}
The given sum telescopes to 3