Could someone tell me how to solve this?
I would use the formulas:
$\displaystyle \sum_{k=1}^n(1)=n$
$\displaystyle \sum_{k=1}^n(k)=\frac{n(n+1)}{2}$
$\displaystyle \sum_{k=1}^n(k^2)=\frac{n(n+1)(2n+1)}{6}$
$\displaystyle \sum_{k=1}^n(k^3)=\frac{n^2(n+1)^2}{4}$
and use the property $\displaystyle \sum_{k=1}^n(a\cdot f(k)\pm b\cdot g(k))=a\cdot\sum_{k=1}^n(f(k))\pm b\cdot\sum_{k=1}^n(g(k))$ to rewrite the sum first, then use the above formulas.
$\displaystyle \sum_{k=1}^{10}(k^3)-2\sum_{k=1}^{10}(k^2)+5\sum_{k = 1}^{10}(k)-\sum_{k = 1}^{10}(1)=$
$\displaystyle \frac{10^2\cdot11^2}{4}-2\cdot\frac{10\cdot11\cdot21}{6}+5\cdot\frac{10 \cdot11}{2}-10=$
$\displaystyle 3025-770+275-10=2520$
You really need to have a live tutor.
Use the above:
$\displaystyle \sum\limits_{k = 1}^{10} {k^3 } - 2\sum\limits_{k = 1}^{10} {k^2 } + 5\sum\limits_{k = 1}^{10} k - \sum\limits_{k = 1}^{10} {(1)} $
Now plug in:
$\displaystyle \frac{10^2(10+1)^2}{4}-2\left(\frac{10(10+1)(2\cdot 10+1)}{6}\right)+5\left( \frac{10(10+1)}{2}\right)- 10 $