Question

Partial Sums Find the first six partial sums $S_1, S_2, S_3, S_4, S_5, S_6$ of the sequence whose nth term is given.

$\frac{1}{3}, \frac{1}{3^2}, \frac{1}{3^3}, \frac{1}{3^4},$….

Short answer

The partial sums are:

$\begin{array}{l}\text{First\hspace{0.17em}\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_1=\frac{1}{3}.\\ \text{Second\hspace{0.17em}\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_2=\frac{4}{9}.\\ \text{Third\hspace{0.17em}\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_3=\frac{13}{27}.\\ \text{Fourth\hspace{0.17em}\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_4=\frac{40}{81}.\\ \text{Fifth\hspace{0.17em}\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_5=\frac{121}{243}.\\ \text{Sixth\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_6=\frac{364}{729}.\end{array}$

Step-by-step solution

Step 1.  Given

The given sequence is $\frac{1}{3}, \frac{1}{3^2}, \frac{1}{3^3}, \frac{1}{3^4},$…

Step 2.  To determine

We have to find the first six partial sums.

Step 3.  Calculation

The given sequence is $\frac{1}{3}, \frac{1}{3^2}, \frac{1}{3^3}, \frac{1}{3^4},$…

We see that the exponents of the denominator are increasing by 1.

So, the first 6 terms of these sequence are $\frac{1}{3}, \frac{1}{3^2}, \frac{1}{3^3}, \frac{1}{3^4}, \frac{1}{3^5}, \frac{1}{3^6}$.

Hence,

$\begin{array}{l}\text{First\hspace{0.17em}\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_1=\frac{1}{3}.\\ \text{Second\hspace{0.17em}\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_2=\frac{1}{3}+\frac{1}{3^2}=\frac{4}{9}.\\ \text{Third\hspace{0.17em}\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_3=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}=\frac{13}{27}.\\ \text{Fourth\hspace{0.17em}\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_4=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}=\frac{40}{81}.\\ \text{Fifth\hspace{0.17em}\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_5=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}=\frac{121}{243}.\\ \text{Sixth\hspace{0.17em}partial\hspace{0.17em}\hspace{0.17em}sum}=S_6=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}=\frac{364}{729}.\end{array}$