I'm guessing S^20 is actually referring to the 20th partial sum [tex]S_{20}[/tex], with
[tex]S_{20}=1+(1+1\times0.1)+(1+2\times0.1)+\cdots+(1+18\times0.1)+(1+19\times0.1)[/tex]
in the arithmetic case, and
[tex]S_{20}=1+1.1+1.1^2+\cdots+1.1^{18}+1.1^{19}[/tex]
in the geometric case.
(a) Combining like terms, we have
[tex]S_{20}=20+0.1(1+2+\cdots+18+19)[/tex]
and invoking the formula
[tex]1+2+\cdots+(n-1)+n=\dfrac{n(n+1)}2[/tex]
we end up with
[tex]S_{20}=20+0.1\dfrac{19\times20}2=39[/tex]
(b) Multiply [tex]S_{20}[/tex] by 1.1, then subtract this from [tex]S_{20}[/tex]:
[tex]1.1S_{20}=1.1+1.1^2+1.1^3+\cdots+1.1^{19}+1.1^{20}[/tex]
[tex]S_{20}-1.1S_{20}=-0.1S_{20}=1-1.1^{20}[/tex]
[tex]\implies S_{20}=\dfrac{1.1^{20}-1}{0.1}\approx57.275[/tex]
