\(6,=13\left(42+2\times11+3\times12\right)=13\times100=1300\\ 7,=\left(2017-1017\right)\left(2017+1017\right)=1000\times3034=3034000\\ 8,=\dfrac{\left(2017+1\right)\left[\left(2017-1\right):2+1\right]}{2}\\ =\dfrac{2018\times1009}{2}=1018081\\ 10,=100020001\)