How many terms of the geometric sequence $2, 8, 32, 128,\dots $are required to give a sum of $174,762$?

Question

How many terms of the geometric sequence $2, 8, 32, 128,\dots $are required to give a sum of $174,762$?

My attempt

$a = 2$ (the first term)
$r = 4$ (the common ratio)
$S_n = 174,762$ (sum to $n$ terms)

Using the formula

$$S_n=a\bigg(\frac{r^n-1}{r-1}\bigg)$$

Therefore

\begin{align*} S_n&=2\bigg(\frac{4^n-1}{4-1}\bigg)\\[5pt] S_n&=174,762\\[5pt] 4^n-1&= \frac{174,762\cdot3}{2} \\[5pt] 4^n &= 262,144 \\[5pt] n &= \log_4(262,144) \\[5pt] n & =\frac{\ln(262,144)}{\ln(4)}\\[5pt] n &= \frac{12.476...}{1.386...} \\[5pt] n & = 9 \end{align*}

There 9 terms are required. Is there a better way go about this? Is this process succinct?

Answer 1

Once you reach $$ 4^n = 262,144 $$ I would proceed with arithmetic, not logarithms. You know $$ 4^5 = 2^{10} = 1,024 $$ so $$ 4^{10} = 2^{20} \approx 1,000,000 $$ is nearly four times too big. So $4^9 = 262,144$ and you need $9$ terms.

Answer 2

That is correct and it is a reasonable way to solve the problem. Of course you may add up the terms until you get to the answer due to fast growth of the terms.