15 Answer: S₁ = 1 S₂ = 4 S₃ = 13 S₄ = 40 Sum = NO
Step-by-step explanation:
1 + 3 + 9 + 27 + ... [tex]\implies\sum^{\infty}_{n=1}3^{n-1}\implies\sum^{\infty}_{n=1}\dfrac{3^n}{3}\\\\\bullet S_1=1\\\bullet S_2=1+3=4\\\bullet S_3=1+3+9=13\\\bullet S=1+3+9+27=40\\\\\\ \lim_{n \to \infty} \dfrac{3^n}{3} \implies\dfrac{3^{\infty}}{3}\implies\infty\\\\\text{The series diverges so there is no sum.}[/tex]
16 Answer: [tex]\bold{S_1=\dfrac{1}{2}\qquad S_2=\dfrac{2}{3}\qquad S_3=\dfrac{13}{18}\qquad S_4=\dfrac{39}{54}\qquad Sum=YES}[/tex]
Step-by-step explanation:
[tex]\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{162}+...\implies \sum^{\infty}_{n=1}\dfrac{1}{2}\bigg(\dfrac{1}{3}\bigg)^{n-1}\\\\\\\bullet S_1=\dfrac{1}{2}\\\\\bullet S_2=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{2}{3}\\\\\bullet S_3=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{18}=\dfrac{13}{18}\\\\\bullet S_4=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{18}+\dfrac{1}{54}=\dfrac{39}{54}[/tex]
[tex]\lim_{n \to \infty} \dfrac{1}{2}\bigg(\dfrac{1}{3}\bigg)^{n-1}\implies \dfrac{1}{2}\lim_{n \to \infty} \dfrac{1}{3^{\infty-1}}\implies \dfrac{1}{\infty}=0\\\\\\\text{The series converges so it does have a sum.}[/tex]
