Question

Find the sum of series $2 + 6 + 18 + ... + 4374$

Answer 1

Hint: The series of a sequence is the sum of the sequence to a certain number of terms. Here the difference between the numbers is not the same. So, this is a G.P series. We will find a common ratio then we will calculate the number of terms. Further we will use the formula of sum of series to get the answer.
The using formula sum of series, ${S_n} = \dfrac{{a({r^n} - 1)}}{{(r - 1)}}$


Complete step by step solution:
The series is $2 + 6 + 18 + .... + 4374$
Above sequence is Geometric progression with common ratio $ = 3$
$\therefore r = 3\,\,and\,\,a = 2$
Last term $(l) = 4374$
$l({a_n}) = a{r^{n - 1}}$
${a_n} = a{r^{n - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(i)$
Now, putting the value of $a\,\,and\,\,r$in equation $(i)$
$ \Rightarrow 4374 = 2{(3)^{n - 1}}$
$ \Rightarrow \dfrac{{4374}}{2} = {3^n}{.3^{ - 1}}$
$ \Rightarrow \dfrac{{4374}}{2} = {3^n} \times \dfrac{1}{3}$
$ \Rightarrow \dfrac{{4374}}{2} \times \dfrac{3}{1} = {3^n}$
$ \Rightarrow 2187 \times 3 = {3^n}$
We factorize the value of $2187$

$3$	$2187$
$3$	$729$
$3$	$243$
$3$	$81$
$3$	$27$
$3$	$9$
$3$	$3$
	$1$

\[2187 = 3 \times 3 \times 3 \times 3 \times 3 \times 3\]
$2187 = {3^7}$
$\therefore {3^7} \times {3^1} = {3^n}$
$ \Rightarrow {3^{7 + 1}} = {3^n}$
$ \Rightarrow {3^8} = {3^n}$
$\therefore n = 8$
Now, we solve for sum of series
${S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$
$ \Rightarrow {S_n} = \dfrac{{2({3^8} - 1)}}{{3 - 1}}$
${S_n} = \dfrac{{2({3^8} - 1)}}{2}$
${S_n} = {3^8} - 1$
As ${3^8} = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 6561$
$ = 6561 - 1$
$ = 6560$
Hence, the sum of the series is $6560$


Note: Students should check carefully the series whether sometimes it is an AP, sometimes it is in GP etc. Students must know that if the base of the value is the same only then we can equate their powers.