Consecutive integers sum with different steps

This a question of notation.

$1+2+3+4+\dots+n$ is a notation for $\sum_{k=1}^n k$

I assume that $11+22+33+44+\dots+11\times n$ is a notation for $\sum_{k=1}^n 11\times k$

in this case, you just get : $$ \sum_{k=1}^n 11\times k = 11 \times \sum_{k=1}^n k = 11 \frac{n(n+1)}{2}$$

with any $M=3$ or $11$, you get $\sum_{k=1}^n M\times k= M \times \sum_{k=1}^n k$

The general formula derives from the simple one.

The general sequence is $$a=a-b+b,a+b=a-b+2b,a+2b,\cdots a-b+nb,$$ i.e. $n$ terms from $a$ to $a+(n-1)b=a+c$.

Then

$$\sum_{k=1}^n(a-b+kb)=n(a-b)+\frac{n(n+1)}2b=n\frac{a+a+(n-1)b}2=n\frac{a+c}2.$$

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With $11$ to $11n$ in steps $11$,

$$n\frac{11+11n}2=11\frac{n(1+n)}2.$$

You get what is called an arithmetic series. More details are at https://en.wikipedia.org/wiki/Arithmetic_progression .