1
$\begingroup$

I got this math "riddle" in one of my math test, and I would love to know how to solve it.

If $$S = 1 + 2 + 3 + 4 + \ldots + 2015,$$ then a sum of $$1 + 2 + 3 + \ldots + 2015 + 2016 + \ldots + 4030$$ is equal to: ?

I know the answer is $2S + 2015^2$, but not how to get it. Could someone explain, please?

$\endgroup$
  • 1
    $\begingroup$ For the terms 2016 and up, write them as $2015+1, 2015+2, 2015+3, \dots$. $\endgroup$ – GEdgar Apr 10 '15 at 20:15
  • $\begingroup$ A related question would be to evaluate $2015+201S$ :) $\endgroup$ – hypergeometric Apr 11 '15 at 6:16
5
$\begingroup$

$$ \begin{align} &1+2+\cdots+2015+2016+\cdots+4030\\ =\:&S+(2015+1)+(2015+2)+\cdots+(2015+2015)\\ =\:&S+(1+2+\cdots+2015)+(\underbrace{2015+2015+\cdots+2015}_{2015\text{ times}})\\ =\:&S+S+2015\cdot2015\\ =\:&2S+2015^2 \end{align} $$ If you want to find the value of $S$, you can do the following additional pairing: $$ \begin{align} S&=1+2+\cdots+2015\\ &=(2015+1)+(2014+2)+\cdots+(1009+1007)+1008\\ &=\underbrace{2016+2016+\cdots+2016}_{1007\text{ times}}+1008\\ &=1007\cdot2016+1008\\ &=2\,031\,120 \end{align} $$

$\endgroup$
  • $\begingroup$ I see, quite easy when I see it like this.. thanks a lot! :) $\endgroup$ – Mykybo Apr 10 '15 at 20:24
1
$\begingroup$

Let $S$ be the sum $S=\sum_{n=1}^{2015}\, n$. Then, we have

$$\begin{align} \sum_{n=1}^{4030} \,n&=\sum_{n=1}^{2015} \,n+\sum_{n=2016}^{4030} n\\\\ &=\sum_{n=1}^{2015} \,n+\sum_{n=1}^{2015}\, (n+2015)\\\\ &=\sum_{n=1}^{2015} \,n+\sum_{n=1}^{2015}\, n+2015\sum_{n=1}^{2015}\, 1\\\\ &=S+S+(2015)\times (2015)\\\\ &=2S+2015^2 \end{align}$$

As was to be shown.

$\endgroup$

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.