What must be the simplest proof of the sum of first $n$ natural numbers? [duplicate]

I was studying sequence and series and used the formula many times $$1+2+3+\cdots +n=\frac{n(n+1)}{2}$$ I want its proof.

Thanks for any help.

marked as duplicate by Noble Mushtak, user223391, E. Joseph, Adam Hughes, I am BackJan 8 '17 at 18:32

• There exist a similar question btw..., or maybe it is not similar but it is exactly same – I am Back Jan 8 '17 at 18:25

Let the sum be $$S_n=1+2+3+\cdots +n\tag1$$ on reversing the same equation we get $$S_n=n+(n-1)+(n-2)+\cdots +1\tag2$$ On adding $(1)$ and $(2)$ we have each term equal to $n+1$ which will occur $n$ times i.e. $$2S_n=(n+1)+(n+1)+(n+1)\cdots\{n times\}+(n+1)$$ $$2S_n=n(n+1)$$ $\therefore$ $$S_n=\frac{n(n+1)}{2}.$$ Hope it helps!!!

• Nice answer.+1. Thank you for help. – user401699 Jan 10 '17 at 16:35

Hope this helps (think about the area): • Yo..... It is a NT and Geometry combo... +1 – I am Back Jan 8 '17 at 18:33
• @THELONEWOLF. :-P Wait, this is NT? – Simply Beautiful Art Jan 8 '17 at 18:34
• Whatever you say.... Sequence and series or blah blah – I am Back Jan 8 '17 at 18:43

I'm not sure how simple this gets, but I still think it's worth noting that this can be applied to higher powers such as $1^2+2^2+\cdots n^2$ or $1^3+2^3+\cdots n^3$.

Solving by the use of Indeterminate Coefficients:

Assume the series$$1+2+3+4+5\ldots+n\tag1$$ Is equal to the infinite series$$1+2+3+4+5+\ldots+n=A+Bn+Cn^2+Dn^3+En^4+\ldots\&c\tag2$$ If we 'replace' $n$ with $n+1$, we get$$1+2+3+4+\ldots+(n+1)=A+B(n+1)+C(n+1)^2+\ldots\&c\tag3$$ And subtracting $(3)-(2)$, gives\begin{align*} & n+1=B+C(2n+1)\tag4\\n & +1=2Cn+(B+C)\tag5\end{align*} Therefore, $C=\dfrac 12,B=\dfrac 12,A=0$ and $(1)$ becomes$$1+2+3+4+5\ldots+n=\dfrac n2+\dfrac {n^2}2=\dfrac {n(n+1)}{2}$$

• I wish I can vote +1000 ! – Khosrotash Jan 8 '17 at 18:34
• @Khosrotash Lol, thanks! Happy to help! – Frank Jan 8 '17 at 18:36

There are $n$ terms, and the average term is $\dfrac{n+1}2$.

Let $S(n)$ be the statement: $1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$

Basis step: $S(1)$:

LHS $=1$

RHS $=\dfrac{1(1+1)}{2}$

$\hspace{9.25 mm}=\dfrac{2}{2}$

$\hspace{9.25 mm}=1$

$\hspace{47.5 mm}$ LHS $=$ RHS $\hspace{1 mm}$ (verified.)

Inductive step:

Assume $S(k)$ is true, i.e. assume that $1+2+3+\cdots+k=\dfrac{k(k+1)}{2}$

$S(k+1)$: $\underline{1+2+3\cdots+k}+(k+1)$

$\hspace{12 mm}=\dfrac{k(k+1)}{2}+k+1$

$\hspace{12 mm}=\dfrac{k(k+1)+2(k+1)}{2}$

$\hspace{12 mm}=\dfrac{(k+1)(k+2)}{2}$

So $S(k+1)$ is true whenever $S(k)$ is true.

Therefore, $1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$.

Let $S_n$ be the sum, clearly $S_1 = 1$. If $S_n = {1 \over 2} n (n+1)$, then $S_n+(n+1) = (n+1) ({n \over 2}+1) = {1 \over 2} (n+1)(n+2) = S_{n+1}$.