# Represent $50+49+48$ and so on to zero

How do I mathematically represent the operation of adding $50$, for example, to continuous decrements by $1$ until we reach $0$ ? $50+49+48+47+\dots$ until we reach $0$. For example, how do I put this problem in a way that WolframAlpha would understand and solve?

Mathematically, you could write that as $$\sum_{i=0}^{50} (n-i)$$

which is the same as $$\sum_{i=0}^{50} i$$

Wolfram Alpha understands quite a bit of mathematical terms, so you can simply search for

sum of i for i from 0 to 50


http://www.wolframalpha.com/input/?i=sum+of+i+for+i+from+0+to+50

• Sorry...I have mistakenly edited your answers while posting mine....pls forgive me and rolled back the edited... – tatan Jun 1 '16 at 12:21
• @tatan What the hell did you do to my answer?! – 5xum Jun 1 '16 at 12:22
• I am really sorry...please rolled it back... – tatan Jun 1 '16 at 12:23
• @tatan lol no problem, no harm done. – 5xum Jun 1 '16 at 12:23
• Thanks a lot for a well-written and comprehensive answer. – Rok Jun 2 '16 at 7:30

Sum of first $n$ natural numbers is given by

$$S_n=\frac {n (n+1)}{2}$$

Using this you can find your sum without any calculator...

• That's pretty awesome. I didn't know that; thanks! – Rok Jun 2 '16 at 7:27
• Happy to help you!!! – tatan Jun 2 '16 at 7:43

In Wolfram|Alpha we can write:

$$\operatorname{sum}[\operatorname{i},\operatorname{i}=0..50]$$

giving $1275$.

A Wolframalpha syntaxis for $$\sum_{k=a}^b g(k)$$ is

sum( g(k), k, a, b)


And it's even surprisingly versatile symbolically...

Your case is $a=0,\ b=50,\ g(k)=50-k$

Side note: There is a counterpart for $\int_a^b g(x)\,dx$

int(g(x), x, a, b)


In mathematical notation, this would be written as $\sum_{k=0}^{50}k$, but Wolfram|Alpha would rather see it as something like sum of k from 0 to 50, and they also have widgets. Of course, you would always use the formal method (the one on the first line) in a formal paper, proof, or even a forum like here, but if you are just wanting to calculate it, you could send it to Wolfram|Alpha in the form on the second line.