Find 10 partial sums of the series. (round your answers to five decimal places.) ∞ 15 (−4)n n = 1

Given

[tex]\Sigma_{n=1}^\infty15(-4)^n[/tex]

The first 10 partial sums are as follows:

[tex]S_1=\Sigma_{n=1}^{1}15(-4)^n=15(-4)=\bold{-60} \\ \\ S_2=\Sigma_{n=1}^{2}15(-4)^n=\Sigma_{n=1}^{1}15(-4)^n+15(-4)^2 \\ =-60+15(16)=-60+240=\bold{180} \\ \\ S_3=\Sigma_{n=1}^{3}15(-4)^n=\Sigma_{n=1}^{2}15(-4)^n+15(-4)^3 \\ =180+15(-64)=180-960=\bold{-780} \\ \\ S_4=\Sigma_{n=1}^{4}15(-4)^n=\Sigma_{n=1}^{3}15(-4)^n+15(-4)^4 \\ =-780+15(256)=-780+3,840=\bold{3,060} \\ \\ S_5=\Sigma_{n=1}^{5}15(-4)^n=\Sigma_{n=1}^{4}15(-4)^n+15(-4)^5 \\ =3,060+15(-1,024)=3,060-15,360=\bold{-12,300}[/tex]

[tex]S_6=\Sigma_{n=1}^{6}15(-4)^n=\Sigma_{n=1}^{5}15(-4)^n+15(-4)^6 \\ =-12,300+15(4,096)=-12,300+61,440=\bold{49,140}[/tex]

The rest of the partial sums can be obtained in similar way.