<s>[INST] Different ways to come up with $1+2+3+\cdots +n=\frac{n(n+1)}{2}$ I am trying to compile a list of the different ways to come up with the closed form for the sum in the title. So far I have the famous Gauss story, the argument by counting doubletons, and using generating functions. Is there any other (significantly) different way to discover the closed form?
To clarify, Gauss's method was to take $1+2+3+\dots+n$, write it backwards, and add it up.
 [/INST] Here is one with the use of Indeterminant Coefficients:
Starting with $$1+2+3+4+5+\cdots+n\tag1$$Assume it to be equal to $A+Bn+Cn^2+Dn^3+\cdots\&\text c$. Hence$$1+2+3+4+5+\cdots+n=A+Bn+Cn^2+Dn^3+En^4+\cdots\&\text c\tag2$$
Therefore, we also have$$1+2+3+4+\cdots+n+1=A+B(n+1)+C(n+1)^2+D(n+1)^3+\cdots\&\text c\tag3$$
Subtracting $(2)$ from $(3)$, we get$$n+1=B+C(2n+1)\tag4$$And equating coefficients, we have$$2C=1\\B+C=1\\\implies C=B=\dfrac 12,\ A=0$$
Therefore$$1+2+3+4+\cdots n=0+\dfrac 12n+\dfrac 12n^2=\dfrac {n^2+n}2\tag5$$
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