Find the sum $\sum_{j=0}^{n}j$
for all $n=0,1,2,3,\dots$.
How do I find/evaluate this sum?
Asked
Active
Viewed 213 times
4 Answers
4
HINT
- add first and last term to get $n+1$
- next pair (next to last and second) gives the same result
- see Gauss Trick
gt6989b
- 54,422
-
I don't get your first hint, could you elaborate? – jukka.aalto Nov 04 '15 at 14:59
-
@jukka.aalto first term in your sum and last term added together are $n+1$. Next pair is $2 + (n-1) = n+1$. etc – gt6989b Nov 04 '15 at 15:27
3
Consider: $$S = \sum_{j=0}^n x^j = \frac{x^{n+1} - 1}{x - 1}$$ $$\frac {d}{dx}S = \sum_{j=0}^njx^{j-1} = \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2}$$
If we take the limit as $x \to 1$, the left hand side becomes $\sum_{j=0}^n j$, and we can evaluate the RHS to find a function of $n$ $$\lim_{x\to 1}\frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2} = \lim_{x\to1}\frac{n(n+1)x^n - (n+1)nx^{n-1}}{2(x-1)}$$ $$= \lim_{x\to1}\frac{n(n+1)x^n - (n+1)nx^{n-1}}{2(x-1)} = \lim_{x\to1}x^{n-1}\frac{(n+1)n}{2} = \frac{(n+1)n}{2}$$
jameselmore
- 5,207
-
-
@SimonS It was more of a sarcastic answer. In fact I'm surprised no one has flagged this question as a duplicate yet – jameselmore Nov 04 '15 at 15:24
1
Here is a high school argument which I love more! :)
Assume $n$ is an odd positive integer then
$$\eqalign{ & \sum\limits_{j = 0}^n j = 0 + 1 + 2 + ... + (n - 2) + (n - 1) + n \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {0 + n} \right) + \left( {1 + \left( {n - 1} \right)} \right) + \left( {2 + \left( {n - 2} \right)} \right) + ... \cr & \,\,\,\,\,\,\,\,\,\,\, = \underbrace {n + n + n + ...}_{{{n + 1} \over 2}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {{{n + 1} \over 2}} \right)n = {{n(n + 1)} \over 2} \cr} $$
you can argue most similarly when $n$ is an even number! I will leave this case as an exercise for you. :)
Hosein Rahnama
- 14,882