summation of multiplying series with negative terms

Question

i know the summation of $1 \times 2 + 2\times 3 + 3\times 4 = \frac{k(k+1)(k+2)}3$ but this one has negative terms. so we need to find below

$$\sum_{i=1}^n(-1)^{i+1} (i(i+1))$$

e.g. $1\times 2 - 2\times 3 + 3\times 4 ....$

if you can share the derivation then it would be very great.

Answer 1

Hint:

If $n$ is even,

\begin{align} \sum_{i=1}^n (-1)^{i+1} i(i+1) &= 1(2)-(2)(3)+(3)(4)-4(5)+\ldots+(n-1)(n)-n(n+1) \\ &=2(1-3)+4(3-5)+\ldots+n(n-1-n-1)\\ &=(-2)(2+4+\ldots n)\\ &=(-4)\left(1+2 + \ldots +\frac{n}{2}\right)\end{align}

Edit:

If $n$ is even, the expression above simplifies to

$$(-4)\frac12\left(\frac{n}2 \right)\left(\frac{n}2 +1\right)=-\frac{n(n+2)}2$$

If $n$ is odd, then $n-1$ is even and the corresponding expression is

$$-\frac{(n-1)(n+1)}2+n(n+1)=(n+1)\left(n - \frac{n-1}{2} \right)=\frac{(n+1)^2}{2}$$

Answer 2

Note that $\sum_{n=1}^{2k} (-1)^{n} n(n+1) = \sum_{n=1}^{2k} (-1)^{n} n^{2} + \sum_{n=1}^{2k} (-1)^{n} n$

So we now evaluate each of the sum and finally add them up to get the required value!

now $\sum_{n=1}^{2k} (-1)^{n} n^{2} = k(2k+1)$,see here

also if we look a bit carefully to $\sum_{n=1}^{2k} (-1)^{n} n$,the value is $k$ as there will be $k$ pairs summing to 1.

so final value is $\sum_{n=1}^{2k} (-1)^{n} n^{2} = k(2k+1) + k = 2k(k+1)$

Hope this helps!

Answer 3

Let $$S_+(n) = \sum_{i=0}^n i (i+1) = 1\cdot2 + 2\cdot3 + 3\cdot4+\dots+n(n+1)$$ and $$S_-(n) = \sum_{i=0}^n (-1)^i i (i+1) = 1\cdot2 - 2\cdot3 + 3\cdot4-\dots + (-1)^n n(n+1)$$

$\frac{S_+(n) - S_-(n)}{2} = 1\cdot2 + 3\cdot4 + 5\cdot6 \dots$ and can be written as $$ \frac{S_+(n) - S_-(n)}{2} = 2\cdot2 + 4\cdot4 + 6\cdot6 + \dots + n^2 - (2+4+6+\dots+2n), \text{$n$ even} $$ $$ \frac{S_+(n) - S_-(n)}{2} = 2\cdot2 + 4\cdot4 + 6\cdot6 + \dots + (n-1)^2 - (2+4+6+\dots+2(n-1)), \text{$n$ odd} $$ Now, use the fact that $\sum_{i=0}^n i^2 =\frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=0}^n i = \frac{n(n+1)}{2}$.

Now rearrange to get $S_-(n)$.

Answer 4

Just another way to do it.

Consider $$S_n=\sum_{i=0}^n i(i+1)x^{n+1}$$ (notice that, because of the $i$, you can start the summations at $0$ instead of $1$) and write $$i(i+1)x^{i+1}=(i+2)(i+1)x^{i+1}-2ix^{i+1}-2x^{i+1}$$ that is to say $$i(i+1)x^{i+1}=x\left(x^{i+2}\right)''-2x^2\left(x^{i}\right)'-2x\left(x^{i}\right)$$ mking $$S_n=x\left(\sum_{i=0}^n x^{i+2}\right)''-2x^2\left(\sum_{i=0}^n x^{i}\right)'-2x\left(\sum_{i=0}^n x^{i}\right)$$ Now, use $$\sum_{i=0}^n x^{i+2}=\frac{x^2 \left(x^{n+1}-1\right)}{x-1}\qquad \text{and}\qquad \sum_{i=0}^n x^{i}=\frac{x^{n+1}-1}{x-1}$$ Compute the required derivatives to end with $$S_n=???$$ and replace $x$ by $-1$ to get the result.