The first term of a geometric sequence is 3 and the multiplier, or ratio, is 5. What is the sum of the first 4 terms of the sequence?

Apr 30, 2018

${S}_{4} = 468$

Explanation:

If we have a general case, let's say the geometric sequence ${\delta}_{n}$, with first term ${\delta}_{1}$ and ratio $q$.

Then, the $k$-th term of the sequence, for an integer $k$, is:

${\delta}_{k} = {\delta}_{1} \cdot {q}^{k - 1}$

Using this knowledge, let's calculate the sum of the first $n$ terms:

${\delta}_{1} + {\delta}_{2} + {\delta}_{3} + \ldots + {\delta}_{n - 1} + {\delta}_{n} = {S}_{n}$
${\delta}_{1} + {\delta}_{1} q + {\delta}_{1} {q}^{2} + \ldots + {\delta}_{1} {q}^{n - 2} + {\delta}_{1} {q}^{n - 1} = {S}_{n}$

Now, multiply both sides by $q$:

$\textcolor{red}{{\delta}_{1} q + {\delta}_{1} {q}^{2} + \ldots + {\delta}_{1} {q}^{n - 1}} + {\delta}_{1} {q}^{n} = {S}_{n} q$

Notice how the highlighted part is the sum of the first $n$-th terms, without the first term, ${\delta}_{1}$.

Hence, we have:

$\textcolor{red}{{S}_{n} - {\delta}_{1}} + {\delta}_{n} = {S}_{n} q$

Substract ${S}_{n}$ on both sides.

${\delta}_{n} - {\delta}_{1} = {S}_{n} \left(q - 1\right)$
${\delta}_{1} {q}^{n - 1} - {\delta}_{1} = {S}_{n} \left(q - 1\right)$
${\delta}_{1} \left({q}^{n - 1} - 1\right) = {S}_{n} \left(q - 1\right)$

Finally, we have

color(red)(S_n = delta_1 (q^(n-1)-1)/(q-1))="first term"*("ratio"^"number of terms"-1)/("ratio"-1)

In our case, the first term is $3$, the ratio/multiplier is $5$ and we wish to add $4$ terms.

${S}_{4} = 3 \cdot \frac{{5}^{4} - 1}{5 - 1} = 468$