First Problem in 2016

Number Theory Level 4

∑ k = 2 ∞ 0. 01 ‾ k = S 4 \displaystyle\sum_{k=2}^{\infty}0.\overline{01}_{k}=S_{4}

If S = p q S=\dfrac{p}{q} , where p p and q q are coprime positive integers, find p + q p+q .

Note : S 4 S_{4} represents S S written in base 4.


The answer is 13.

View wiki
Arsan Safeen by Arsan Safeen , 23, Iraq

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Shaun Leong
Jan 7, 2016

Let x k = 0.0101 … k x_k=0.0101\ldots_k k 2 x k = 1.0101 … k^2x_k=1.0101\ldots k 2 x k − x k = 1 k^2x_k-x_k=1 x k = 1 ( k − 1 ) ( k + 1 ) = 1 2 ( 1 k − 1 − 1 k + 1 ) x_k=\frac {1}{(k-1)(k+1)}=\frac {1}{2} (\frac {1}{k-1}-\frac {1}{k+1})

Hence ∑ k = 2 ∞ x k = 1 2 ( 1 1 − 1 3 + 1 2 − 1 4 + 1 3 − 1 5 + … ) \displaystyle \sum_{k=2}^\infty x_k=\frac {1}{2}(\frac {1}{1}-\frac {1}{3}+ \frac {1}{2}- \frac {1}{4} +\frac {1}{3}- \frac {1}{5}+\ldots) = 1 2 ( 1 1 + 1 2 ) = 3 4 =\frac {1}{2}(\frac {1}{1} +\frac {1}{2})=\frac {3}{4} = 3 ( 4 − 1 ) = 0. 3 4 =3 (4^{-1})=0.3_4

Thus S = 3 10 ⇒ 3 + 10 = 13 S=\frac {3}{10} \Rightarrow 3+10=\boxed {13}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...