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I have a series in my own:

$1+(1+2)+(1+2+3)+...+(1+...+10)$

How I can show these with Sigma notation and Reach to 220?

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$$\sum_{m=1}^n(1+2+\cdots+m)=\sum_{m=1}^n\sum_{k=1}^mk$$ Also we can obtain a closed formula as $$\dfrac{n}{6}(n+1)(n+2)$$ for this summation for any $n\in\mathbb{N}.$

Added:
There are multiple ways to establish this general formula. You can get some proof ideas from here, which discusses many different proofs of the fact that $$\color{Green}{1+2+\cdots+n=\dfrac{n}{2}(n+1)}.$$ Since there are many nice internet sources on general techniques to sum (finite) series, I will give you a sort of intuitive explanation using only the formula in green. $$S=1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+n)\tag1$$ First, we can simplify the first equation using the green formula to obtain $$2S=\color{Blue}{2}+\color{Red}{6}+\cdots+\color{Orange}{n(n-1)}+n(n+1)\tag2.$$ Secondly, notice that in the first series one appear $n$ times and two appear $n-1$ times and so on. Therefore we can rearrange the series as $$S=1.n+\color{Blue}{2.(n-1)}+\color{Red}{3.(n-2)}+\cdots+\color{Orange}{n.1}\tag3$$ Now, lets add second and third equation according to given color codes to get $$3S=n+\color{Blue}{2n}+\color{Red}{3n}+\cdots+\color{Orange}{n^2}+n(n+1)\tag4.$$ Finally factor $n$ from the right hand side of this fourth equation and apply the formula in green once again for $n+1$ terms :)

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You have $10$ sums, so you want to start with $$\sum_{k=1}^{10}$$ Sum number $k$ is of the form $$1+2+\cdots+k=\sum_{i=1}^ki$$ Altogether, we have $$\sum_{k=1}^{10}\sum_{i=1}^ki$$ As for evaluating the sum, use the formula for the sum of an arithmetic progression for the inner sums. Then you have $10$ numbers to add up.

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