Finding partial sum of a series

Question

I need to find sum of n terms of series: $$-1 +2-3+4-5+6 .... $$ Which is in the form $$\sum_{k=1}^{n}(-1)^{k}k$$ I am not sure how i should approach this

Answer 1

Let $S_n=\sum_{k=0}^nr^k$. This is famously known as the geometric series:

$$S_n(r)=\sum_{k=0}^nr^k=\frac{1-r^{n+1}}{1-r}$$

If we take the derivative of both sides, we end up with

$$S_n'(r)=\sum_{k=1}^nkr^{k-1}=\frac d{dr}\frac{1-r^{n+1}}{1-r}$$

Let $r=-1$ and multiply both sides by $-1$ to get

$$\sum_{k=1}^nk(-1)^k=-\left(\frac d{dr}\frac{1-r^{n+1}}{1-r}\bigg|_{r=-1}\right)$$

Answer 2

HINT

Notice $$ (-1+2) + (-3+4) + \ldots = 1 + 1 + \ldots $$

Answer 3

Work out your partial sums.

$S_1 = -1 $

$S_2 = -1 + 2 = 1$

$S_3 = S_2 - 3 = -2$

$S_4 = S_2 + 4 = 2$

$S_5 = S_4 -5 = -3$

$S_6 = S_2 + 6 = 3$

Clearly this pattern continues.

You can find two separate expressions, one for n even and one for n odd,

OR you can use the "floor" or "ceiling" function

$\lfloor{(n-1)/2} \rfloor$ = greatest integer $\leq (n-1)/2$, or

or $\lceil (n/2) \rceil = $ least integer $\geq (n/2)$

With one of these expressions and $(-1)^n$, construct your formula for $S_n$.

Answer 4

When $n$ is even, such as $$-1+2-3+4-5+6=\color{red}{-1+1}+1\color{red}{-3+3}+1\color{red}{-5+5}+1=3$$ So inductively the total is $$\frac n2$$

When $n$ is odd, such as $$-1+2-3+4-5+6-7=\color{red}{-1+1}+1\color{red}{-3+3}+1\color{red}{-5+5}+1-7=-4$$ So inductively the total is $$-\frac{n+1}{2}$$

Answer 5

Let $s(n)=-1 +2-3+4-5+6 ....+(-1)^nn = \sum_{k=1}^n(-1)^kn$

We now have that $$s(n+1)=\sum_{k=1}^{n+1}(-1)^k(n+1)$$ $$s(n+1)=(-1)^{n+1}(n+1)+\sum_{k=1}^{n}(-1)^kn+\sum_{k=1}^{n}(-1)^k$$ $$s(n+1)=(-1)^{n+1}(n+1)+s(n)+\frac{1}{2}[(-1)^n - 1]$$
We can subtract $s(n)$ from both sides and get a finite difference equation which we can happily solve with any common technique