1) We have to find the sum of the first four terms of the geometric sequence:
[tex]3+3(\frac{1}{4})+3(\frac{1}{4})^2+3(\frac{1}{3})^3+3(\frac{1}{4})^4[/tex]
In this case, we can take out the factor 3 and we have a common ratio r = 1/4. We have to add the first 5 terms.
Then, the sum can be expressed as:
[tex]S_n=\frac{a_1(1-r^n)}{1-r}[/tex]
For this problem, r1 = 3, r = 1/4 and n = 5:
[tex]\begin{gathered} S_5=\frac{3(1-(\frac{1}{4})^5)}{1-\frac{1}{4}} \\ S_5=\frac{3(1-\frac{1}{1024}_)}{\frac{3}{4}} \\ S_5=\frac{3(\frac{1024-1}{1024})}{\frac{3}{4}} \\ S_5=\frac{4}{3}\cdot3\cdot\frac{1023}{1024} \\ S_5=\frac{1023}{256} \end{gathered}[/tex]
2) We have this sum already solved but we can check it as:
[tex]\begin{gathered} S=\sum_{i\mathop{=}1}^7(-3)^i \\ S=(-3)+(-3)^2+(-3)^3+(-3)^4+(-3)^5+(-3)^6+(-3)^7 \\ S=-3+9-27+81-243+729-2187 \\ S=-1641 \end{gathered}[/tex]
Answer:
1) 1023/256
2) -1641
