Question

Arrange the geometric series from least to greatest based on the value of their sums

Answer 1

Answer:

$\sum_{i}^{4}2(3)^{i-1}<\sum_{i}^{5}3(2)^{i-1}<\sum_{i}^{7}(2)^{i-1}<\sum_{i}^{5}(3)^{i-1}$.

Step-by-step explanation:

The sum of finite geometric series is given as:

$S_{n}=\sum_{i}^{n}a_{i}r^{i-1}=a_{1}(\frac{1-r^n}{1-r})$

Now, $\sum_{i}^{4}2(3)^{i-1}$ can be solved as:

$\sum_{i}^{4}2(3)^{i-1}=2(\frac{1-3^4}{1-3})=2(\frac{1-81}{-2})=80$

Also, $\sum_{i}^{5}3(2)^{i-1}$ can be solved as:

$\sum_{i}^{5}3(2)^{i-1}=3(\frac{1-2^5}{1-2})=3(\frac{1-32}{-1})=93$

Also, $\sum_{i}^{7}(2)^{i-1}$can be solved as:

$\sum_{i}^{7}(2)^{i-1}=\frac{1-2^7}{1-2}=\frac{1-128}{-1}=127$

and $\sum_{i}^{5}(3)^{i-1}$ can be solved as:

$\sum_{i}^{5}(3)^{i-1}=\frac{1-3^7}{1-3}=\frac{1-2187}{-2}=1093$

We can observe that $80<93<127<1093$, thus

$\sum_{i}^{4}2(3)^{i-1}<\sum_{i}^{5}3(2)^{i-1}<\sum_{i}^{7}(2)^{i-1}<\sum_{i}^{5}(3)^{i-1}$.

Answer 2

Answer:

The above answer is wrong. The last two boxes should be switched.