Series question with unknown limit I am doing an A Level Core 2 paper and the last question caught me out. The question is:
$$\sum_1^{2N} (3r+5) - \sum_1^N (3r+5) = 1256$$
Find N.

How can I do this? I've worked out so far that the LHS can be written as $\sum_N^{2N} 3r+5$ but I can't figure out where to go from there. How do I find N? 

Hints:
$$\sum_{n=1}^N3n=3\sum_{n=1}^Nn=3\frac{N(N+1)}2$$
Since you actually have, say
$$\sum_{r=1}^N (3r+5)=3\sum_{r=1}^Nr+\sum_{r=1}^N5=3\frac{N(N+1)}2+5N$$
Proof of the first sum above: we do the same sum in two ways
$$\begin{align}&S=1&+&2&+\ldots+&(N-1)&+&N\\
&S=N&+&(N-1)&+\ldots+&2&+&1\end{align}$$
Sum now both lines above:
$$2S=\overbrace{(N+1)+(N+1)+\ldots+(N+1)}^{n\;\text{ times}}=N(N+1)$$
and now you can deduce the sum of the first $\;n\;$ consecutive natural numbers.
