400 Followers Problem

write as: $\dfrac{1}{2}\left(\dfrac{2^n}{1+2^n+2^{n+1}(1+2^n)}\right)$ $=\dfrac{1}{2}\left(\dfrac{2^{n+1}+1-(2^n+1)}{(1+2^n)(1+2^{n+1})}\right)$ $=\dfrac{1}{2}\left(\dfrac{1}{2^n+1}-\dfrac{1}{2^{n+1}+1}\right)$ so the summation becomes: $\sum_{n=1}^{10} \left(\dfrac{1}{2}\left(\dfrac{1}{2^n+1}-\dfrac{1}{2^{n+1}+1}\right)\right)$ this easily telescopes to $\dfrac{1}{2}\left(\dfrac{1}{2^1+1}-\dfrac{1}{2^{11}+1}\right)=\dfrac{341}{2049}$ $341+2049=\boxed{2390}$

Notice that the denominator can be factored as

$(2^{n+1}+1)(2^{n}+1)$

I have a lemma, which states that

$\frac{2^{n-1}}{(2^{n+1}+1)(2^{n}+1)}$

$= \frac{1}{2(2^{n}+1)} - \frac{1}{2(2^{n+1}+1)}$

We can prove this by creating a common denominator, and then dividing both the numerator and denominator by 4.

Then, the summation telescopes to become

$\frac{1}{2(2^{1}+1)} - \frac{1}{2(2^{11}+1)}$

which simplifies to $\frac{341}{2049}$ => $\boxed{2390}$