I want to have a short way to find the sum of $(1+2+3+4+5+6+7+8+9+10+11)$, so is here any short way to sum these numbers? thanks in advance
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2These are triangle numbers. – Simply Beautiful Art Aug 17 '17 at 20:24
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If you don't know the general "Gaussian" trick, you might nonetheless notice that lots of pairs sum to $10$, i.e., $(1+9)+(2+8)+(3+7)+(4+6)+5+10+11=40+5+10+11=66$. – Barry Cipra Aug 18 '17 at 19:14
2 Answers
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There is a very famous identity.
$$1+2+3+...+n=\frac{n(n+1)}{2}$$
Your case would be $n=11$.
As a simple proof, you can consider twice the value, and group the terms:
$$2[1+2+...+(n-1)+n]=\begin{matrix} &[1&+&2&+&...&+&(n-1)&+&n]\\ +&[1&+&2&+&...&+&(n-1)&+&n] \end{matrix}$$
Which we can rearrange.
$$2[1+2+...+(n-1)+n]=\begin{matrix} &[1&+&2&+&...&+&(n-1)&+&n]\\ +&[n&+&(n-1)&+&...&+&2&+&1] \end{matrix}$$
Now, we can add elements vertically, noticing that each sum is $n+1$
$$2[1+2+...+(n-1)+n]=[(n+1)+(n+1)+...+(n+1)]$$
since there are n terms:
$$2[1+2+...+(n-1)+n]=n(n+1)$$
$$1+2+...+(n-1)+n=\frac{n(n+1)}2$$
Kajelad
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What do you mean by shorter?! Add $n$ and its square, then multiply by a half. Two multiplications and an addition. What could be shorter? – Xander Henderson Aug 17 '17 at 19:00
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so what about this formula ( middle number * last number = answer ). (6*11=66) – Ziaurhman Zia Aug 17 '17 at 19:04
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@ZiaurhmanZia When $n$ is even, the "middle number" is not an integer. E.g., $1+2+3+4=\frac{4\cdot 5}{2}=10$ and $2.5\cdot 4=10$. – user236182 Aug 17 '17 at 19:31
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One time write the numbers by the usuall order and one time in the reverse order; and then add them together :
$$\underbrace {\begin {array} {ccccccccccc} 1&2&3&4&5&6&7&8&9&10&11 \\ 11&10&9&8&7&6&5&4&3&2&1 \\ \hline 12&12&12&12&12&12&12&12&12&12&12& \\ \end {array}}_{11-times}$$
So "twice your sum" is equal to $11.12$;
So "your sum" is equal to $\dfrac{11.12}{2}$.
Davood
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you''re right but how about this formula ( the middle number * last number = answer ). (6*11=66) – Ziaurhman Zia Aug 17 '17 at 19:15
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@ Ziaurhman Zia See here: https://math.stackexchange.com/questions/2260/proof-for-formula-for-sum-of-sequence-123-ldotsn – Davood Aug 17 '17 at 19:17
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@ZiaurhmanZia When $n$ is even, the "middle number" is not an integer. E.g., $1+2+3+4=\frac{4\cdot 5}{2}=10$ and $2.5\cdot 4=10$. – user236182 Aug 17 '17 at 19:33