Question

Find the sum of the series \[2 + 6 + 18 + ... + 4374\]

Answer 1

Hint: In the problem they have mentioned to find the sum of the given series. The given series is in G.P. which means geometric progression and it is represented in the form $a,ar,a{r^2},...$.

Formula Used:
The sum of first n terms in a G.P . The series is denoted by ${s_n}$. The formula used to find the sum of a G.P. series is given by ${s_n} = a(\dfrac{{{r^n} - 1}}{{r - 1}})$, where $a$ is the first term of the series and $r = (\dfrac{{a{r^{n - 1}}}}{{a{r^{n - 2}}}})$ is the common difference between the two adjacent terms.

Complete step-by-step answer:
According to the given data, \[2 + 6 + 18 + ... + 4374\] is our G.P. series.
Here, the first term \[a = 2\]and to find the common difference between the two adjacent terms, we use the formula $r = (\dfrac{{a{r^{n - 1}}}}{{a{r^{n - 2}}}})$.
Basically ‘$r$’ is the division of the second term to the first term.
Therefore, we have that
$ \Rightarrow r = (\dfrac{{a{r^{n - 1}}}}{{a{r^{n - 2}}}})$
$ \Rightarrow r = \dfrac{6}{2}$$ = 3$
Thereafter, on dividing the terms we get $r = 3$.
Now, according to the given information, we have a G.P. with common ratio as $3$.
$\therefore r = 3$and $a = 2$
Last term $ = $ 4374 $ = a{r^{n - 1}}$

Now, we will apply the formula for finding the sum of a G.P. series which is given by,

${s_n} = a(\dfrac{{{r^n} - 1}}{{r - 1}})$
$ \Rightarrow {s_n} = \dfrac{{a{r^n} - a}}{{r - 1}}$

$ \Rightarrow {s_n} = \dfrac{{a{r^{n - 1}} \times (r - 1)}}{{r - 1}}$

Substituting all the values in the above given expression we get,

$ \Rightarrow {s_n} = \dfrac{{4371 \times (3 - 1)}}{{3 - 1}}$

$ \Rightarrow {s_n} = \dfrac{{13120}}{2}$

Finally we get the sum of the given series to be, ${s_n} = 6560$.

Note: In order to solve problems of this type the key step is to have an understanding of the number system and particularly sequence and series. ${s_n}$. The formula used to find the sum of a G.P. series is given by ${s_n} = a(\dfrac{{{r^n} - 1}}{{r - 1}})$, where $a$ is the first term of the series and $r = (\dfrac{{a{r^{n - 1}}}}{{a{r^{n - 2}}}})$ is the common difference between the two adjacent terms.
Also, this formula is only valid if the value of $r$is equal for every division of two adjacent terms.