How to derive the closed form of the summation $\sum_{i=1}^n d (i - 1)$?

Question

I want to derive the closed form of

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

d is a constant

Don't be harsh on me as I am getting back to math after some years of not using it ;)

I am finding only the result online, but for learning purpose I would love a step by step proof.

Meanwhile I am reading for a while finite calculus as in the end I want not only to understand, but do this myself.

Thank you!

Answer 1

we want: $$S=\sum_{r=1}^n d(r-1)$$ and we know that this is a summation where each term of $(r-1)$ is being multiplied by $d$, so we can say that this is the same as the whole summation multiplied by d, hence: $$\sum_{r=1}^n d(r-1)=d\sum_{r=1}^n (r-1)$$ now we can split this into two different summations: $$d\sum_{r=1}^n (r-1)=d\left(\sum_{r=1}^n r-\sum_{r=1}^n 1\right)$$ There is a known general formula for the summation of r: $$\sum_{r=1}^n r=\frac{n(n+1)}{2}$$ and we can see than $n$ lots of $1$ is just $n$, so: $$\sum_{r=1}^n1=n$$ so our summation becomes: $$S=d\left(\frac{n(n+1)}{2}-n\right)=\frac{dn(n-1)}{2}$$

Answer 2

Hint: It holds $$ \sum_{i=1}^n d(i-1) = d \sum_{i=1}^n (i-1) = d\sum_{i=0}^{n-1} i = d\sum_{i=1}^{n-1} i.$$ (Why?) The final task is to calculate the sum $\sum_{i=1}^{n-1} i$. This can be done by a reordering, see for example here.

Answer 3

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

We can then observe that the summation is just 0 + 1 + 2 + ... + (n-1), this will add up to $\frac{n(n-1)}{2}$. Hence, final answer should be equals to $\frac{dn(n-1)}{2}$.

Answer 4

$S:= d \sum _{i=1}^{n} (i-1);$

$S = d[ 1+2+3+....(n-1)]$; or

in reverse order:

$S=d[(n-1)+(n-2)+..+2+1]$.

Add term by term:

$2S =$

$d [(1+(n-1))+ (2+(n-2))+..$

$...+((n-1)+1)];$

There are $n-1$ summands, each equals $n$.

Hence

$2S= $

$d[n +n +....n+n] = d[(n-1)n];$

$S=(1/2)d(n-1)n$.

Answer 5

Thank you all for the insights, each provides a hint to what I want to learn !

The link below explains how Gauss derived the closed form of the summation from: $$S_N = \sum_{i=1}i$$ to $$S_N = N(N + 1) / 2$$

http://mathandmultimedia.com/2010/09/15/sum-first-n-positive-integers/

I also found this useful (Triangular numbers):

https://en.wikipedia.org/wiki/Triangular_number

Answer 6

The highest power of $i$ in your sum is $1$, so we can try to fit a $1+1=2$nd degree polynomial in $n$: $$\sum\limits_{i=1}^n d(i-1)=An^2+Bn+C$$ Substituting in $n=1$ we get that: $$0=A+B+C$$ And with $n=2$: $$d=4A+2B+C$$ And with $n=3$: $$3d=9A+3B+C$$ So you have a system of $3$ linear equation, which you can solve: $A=\frac{d}{2}$, $B=-\frac{d}{2}$, $C=0$.
From this, we might have that (we don't know that if it's true for $n>3$) $$\sum\limits_{i=1}^n d(i-1)=\frac{d}{2}n^2-\frac{d}{2}n$$ But you can (try to) prove it with induction.

And it works for higher powers too, for example, you can get a closed form for $$\sum\limits_{i=1}^{n} (i^6-2i^5+93i^2+i)$$ as a $6+1=7$th degree polynomial in $n$: $$\sum\limits_{i=1}^{n} (i^6-2i^5+93i^2+i)=A_0+A_1n+A_2n^2+A_3n^3+A_4n^4+A_5n^5+A_6n^6+A_7n^7$$ But it will be a bit harder to solve the system of $7$ linear equations.