Y by HamstPan38825, Adventure10
The real numbers
and
satisfy
and
for every integer
. If
and
, compute
Proposed by Evan Chen







![\[ \sum_{k=1}^{2013} \left( a_kb_{k-1} - a_{k-1}b_k \right). \]](http://latex.artofproblemsolving.com/7/3/8/738564afc7b5f7ae6e7a8f23025f66504df20400.png)
It's February and we'd love to help you find the right course plan!
\[ \begin {eqnarray*} a_n &=& \dfrac {1}{63} \cdot \sqrt {2n+2} + a_{n-1} \\ b_n &=& \dfrac {1}{96} \cdot \sqrt {2n+2} - b_{n-1} \end {eqnarray*} \]
\[ \begin {eqnarray*} a_n &=& a + \dfrac {1}{63} \cdot \displaystyle\sum_{i=1}^{n} \sqrt {2i+2} \\ b_n &=& (-1)^n \cdot \left[ b + \dfrac {1}{96} \cdot \displaystyle\sum_{i=1}^{n} (-1)^i \sqrt {2i + 2} \right]. \end {eqnarray*} \]Note: as you just saw now, I changed my index names. I do that a few times in this solution by interest/mood/laziness/etc.
\[ \begin {eqnarray*} b &=& a + \dfrac {1}{63} \cdot \displaystyle\sum_{i=1}^{2013} \sqrt {2i+2} \\ a &=& - \left( b + \dfrac {1}{96} \cdot \displaystyle\sum_{i=1}^{2013} (-1)^i \cdot \sqrt {2i+2} \right). \end {eqnarray*} \]Re-arranging, we get:
\[ \begin {eqnarray*} b + a &=& \dfrac {\displaystyle\sum_{i=1}^{2013} (-1)^{i+1} \cdot \sqrt {2i+2}}{96}, \\ b - a &=& \dfrac {\displaystyle\sum_{i=1}^{2013} \sqrt {2i+2}}{63}. \end {eqnarray*} \]I guess I use this somewhere. Eyes on the ball!
\[ \begin {align*} a_k b_{k-1} - a_{k-1} b_k &= \left( \dfrac {1}{63} \sqrt {2k+2} + a_{k-1} \right) \cdot b_{k-1} - a_{k-1} \cdot \left( \dfrac {1}{96} \sqrt {2k+2} - b_{k-1} \right) \\&= \left( \dfrac {b_{k-1}}{63} - \dfrac {a_{k-1}}{96} \right) \cdot \sqrt {2k+2} + 2a_{k-1} b_{k-1}. \end {align*} \]Okay. Now, we multiply the two recursive definitions of
\[ \begin {align*} A &= \displaystyle\sum_{k=1}^{2013} (a_kb_{k-1} - a_{k-1}b_k) \\&= \displaystyle\sum_{k=1}^{2013} \left( \dfrac {b_{k-1}}{63} - \dfrac {a_{k-1}}{96} \right) \cdot \sqrt {2k+2} + 2 \cdot \displaystyle\sum_{k=0}^{2012} a_k b_k. \]But, also, note that, since
\[ \begin {align*} \displaystyle\sum_{k=0}^{2012} a_n b_n &= \left( \displaystyle\sum_{k=1}^{2013} a_k b_k \right) - a_{2013} b_{2013} + a_0 b_0 \\&= \displaystyle\sum_{k=1}^{2013} a_k b_k. \]Now, I may have said this before but I'm not quite sure, so I'll say it here. We have:
\[ \begin {align*} A &= \displaystyle\sum_{k=1}^{2013} \left( \dfrac {2k+2}{63 \cdot 96} \right) \\&= \dfrac {2}{63 \cdot 96} \cdot \displaystyle\sum_{k=1}^{2013} (k+1) \\&= \boxed {671}. \]
Something appears to not have loaded correctly.