1
$\begingroup$

I was given a recurrence relation of this series: $$a_{n+1} = k -a_n$$ $S_n$ is the sum of $n$ first terms of the series. So I was given that $$S_{101} = 353$$ $$S_{199} = 696$$ With this information I need to find $a_1$ and $k$.


I found that if $$a_{n+1} = k -a_n$$ Then \begin{align} \ a_{n+2} & = k - a_{n+1} \\ & = k - (k-a_n) \\ \end{align} So $$a_{n+2} = a_n$$

I thought that I could find sums by thinking that the series above is an arithmetic progression, $d = 0$ but it didn't help. Later, I thought that it would be a geometric series, the ratio being $r=1$.

The sum of geometric series would be: $$S_n = \frac{a_1(r^n-1)}{r-1}$$ Thus, $r$ can't be $1$.

Any help?

$\endgroup$
  • $\begingroup$ You are right, except that $r=-1$ ! $\endgroup$ – Yves Daoust Feb 16 '16 at 17:27
  • $\begingroup$ @YvesDaoust Why? $\endgroup$ – Pichi Wuana Feb 16 '16 at 17:34
  • $\begingroup$ How did you conclude that $r=1$ ? $\endgroup$ – Yves Daoust Feb 16 '16 at 17:44
  • $\begingroup$ @YvesDaoust because if it's always the same number then $a_1 * 1 = a_2$ $\endgroup$ – Pichi Wuana Feb 16 '16 at 17:50
  • $\begingroup$ Nope, $a_1\ne a_2$. What you have is $a_1=a_3$. $\endgroup$ – Yves Daoust Feb 16 '16 at 17:53
1
$\begingroup$

You have reached the conclusion that $a_n=a_{n+2}$.

Now all you have left is to solve the following system of two equations in two variables:

  • $51a_1+50a_2=353$
  • $100a_1+99a_2=696$

The solution is $a_1=3$ and $a_2=4$.

And then of course, $k=a_1+a_2=7$.

$\endgroup$
  • $\begingroup$ Oh... right! I thought I would need sum equations. Thanks. $\endgroup$ – Pichi Wuana Feb 16 '16 at 17:19
  • $\begingroup$ @PichiWuana: You're welcome :) $\endgroup$ – barak manos Feb 16 '16 at 17:19
1
$\begingroup$

Let's just unroll the recurrence relation.

$ a_2 = k - a_1 $

$ a_3 = k - a_2 = k - (k - a_1) = a_1 $

$ a_4 = k - a_3 = k - a_1 = a_2 $

$ \cdots $

What do you find?

All odd terms in the sequence is simply $ a_1 $

All even terms in the sequence is simply $ a_2 $

The rest is trivial.

$ \sum\limits_{i = 1}^{101}a_i = \sum\limits_{i = 1}^{51}a_{2i - 1} + \sum\limits_{i = 1}^{50}a_{2i} = 51a_1 + 50a_2 = 353 $

$ \sum\limits_{i = 1}^{199}a_i = \sum\limits_{i = 1}^{100}a_{2i - 1} + \sum\limits_{i = 1}^{99}a_{2i} = 100a_1 + 99a_2 = 696 $

Solving, we get $ a_1 = 3 $ and $ a_2 = 4 $, so $ a_2 = k - a_1 \implies 4 = k - 3 \implies k = 7 $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.