Given the sequence: $2,-6,12,-20,30,-42,... $
How can i find the $S_n$ term of that sequence? (the sum of the $n$ first terms)
I can see that the differences without the sign is an arithmetic progression,
and can get to $a_n$ but can't see how can i get to the $S_n$ term because of that sign.
thanks.
EDIT: im sorry i need to find $s_n$ not as i wrote before.
UPDATE: thanks to @mathlove the solution is:
$S_n=-\frac{n(n+2)}{2}\ \ \ \ \text{($n$ is even)}$
$S_n=\frac{(n+1)^2}{2}\ \ \ \ \text{($n$ is odd)}$.
I'm asking if we can write those two terms as one by some manipulation.
thanks!
$\endgroup$ 23 Answers
$\begingroup$It looks like $a_n=(-1)^n|a_n|$ (alternating positive and negative), so if you know an expression for $|a_n|$ ...
$\endgroup$ $\begingroup$Since $$|a_n|=n(n+1)(\text{why?}),$$ we have $$a_n=(-1)^{n-1}n(n+1).$$
EDIT 2 : If you want $S_n$, you can use the followings : $$a_{2k-1}=2k(2k-1), \ \ a_{2k}=-2k(2k+1)$$ where $k=1,2,3,\cdots.$
For exmaple, if you want $S_{10}$, you'll see $$S_{10}=(a_1+a_3+\cdots +a_9)+(a_2+a_4+\cdots +a_{10})$$ $$=\sum_{k=1}^{5}2k(2k-1)+\sum_{k=1}^{5}\{-2k(2k+1)\}$$ $$=\sum_{k=1}^{5}\{(4k^2-2k)-(4k^2+2k)\}=\sum_{k=1}^{5}(-4k)=-4(1+2+3+4+5).$$
EDIT 2 : We can have an explicit form of $S_n$. $$S_n=-\frac{n(n+2)}{2}\ \ \ \ \text{($n$ is even)}$$ $$S_n=\frac{(n+1)^2}{2}\ \ \ \ \text{($n$ is odd)}.$$
$\endgroup$ 17 $\begingroup$Interpolating (using Newton Interpolation) on the points $b_1 = 2, b_2=6, b_3=12, b_4=20, b_5=30,...$ gives $b_n$ = $n^2 + n$.
Adding in the alternating signs, we get $a_n = (-1)^{n+1}(n^2 + n)$
$\endgroup$
