[10,20] is a Problem!

$(r-10)(r-20) <0$ only for $r \in ]10,20[$ .

Let $\sum_{r=1}^{n} |(r-10)(r-20)| = A_n$ .

We have: $A_{50}=A_{10}+A_{20}-A_{10}+A_{50}-A_{20}$

$A_n=S_n$ only for $S_n>0$ , and $A_n=-S_n$ only for $S_n<0$ i.e. for $r\in ]10,20[$

Hence, $A_{10}=S_{10}$

$[A_{20}-A_{10}]=-[S_{20}-S_{10}]$

$[ A_{50}-A_{20}]=[ S_{50}-S_{20}]$

So, $A_{50}=A_{10}+A_{20}-A_{10}+A_{50}-A_{20}$

$<=> A_{50}=S_{10}-S_{20}+S_{10}+S_{50}-S_{20}$

Therefore, $\boxed{A_{50}=2S_{10}+S_{50}-2S_{20}}$

The difference between $S_{50}$ and the sought for $SUM$ is only that the terms from $r=11$ to $r=11$ flip from negative to posive. We call the (positive) sum of those 9 terms $X$ . We then have:

$SUM=S_{50}+2X$

We notice that $X$ is also the difference between $S_{10}$ and $S_{20}$ :

$S_{20}=S_{10}-X$

Substituting $X$ , we get

$SUM=S_{50}-2S_{20}+2S_{10}$

Notice that $(r-10)(r-20)$ is non-negative except for r in the range from 11 to 19.
Expanding on Hasan's notation, let's let $\sum_{r=m}^n |(r-10)(r-20)| = A_{m,n}$ , with $A_{n} = A_{1,n}.$ Then $S_{10} = A_{10}$ , since the first ten terms of the summation are positive. The next ten terms are non-positive, thus. $S_{n} = A_{10} - A_{11,n}$ , for $11 \leq n \leq 20$ . Finally, as the terms for $n>20$ are all positive, $S_n = A_{10} - A_{11,20} + A_{21,n}$ for $n > 20.$

The question asks for the value of $A_{50}$ in terms of $S_{50}, S_{2}$ and $S_{10}$ .

Notice that $A_{20} = S_{20} + 2*A_{11,20}$ and, indeed, $A_{k}= S_{k} + 2*A_{11,20}$ for $k > 20.$
Also, observe that $S_{10} - S_{20} = A_{11,20}$ . Substitute for $A_{11,20}$ in the equation above to get the result $A_{50} = S_{50} + 2*( S_{10} - S_{20}) = S_{50} + 2*S_{10} - 2*S_{20}$

There is a really simple solution to this.

Arithmetic sum formula.

$\displaystyle S_n = \frac{n(a_1 + a_n)}{2}$

Since the equation is $(r-10)(r-20)$ , this means $a_{10}$ and $a_{20}$ must be $0$ .

Therefore

$S_{10} = S_{20} = \frac{na_1}{2}$

You could solve the for $S_{50}$ and $S_{20}$ but that would be unnecessary.

Just pick the result that add and subtracts the same number and your golden. The one that cancels $S_{20}$ and $S_{10}$