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

My attempt

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

Using the formula

$$S_n=aigg(fracr^n-1r-1igg)$$

Therefore

eginalign*S_n&=2igg(frac4^n-14-1igg)\<5pt>S_n&=174,762\<5pt> 4^n-1&= frac174,762cdot32 \<5pt> 4^n &= 262,144 \<5pt>n &= log_4(262,144) \<5pt>n & =fracln(262,144)ln(4)\<5pt>n &= frac12.476...1.386... \<5pt>n & = 9endalign*

There 9 terms are required. Is over there a much better way go around this? Is this procedure succinct?


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edited Feb 14 in ~ 0:35
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asked Feb 14 at 0:28
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LaufenLaufen
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Once friend reach$$4^n = 262,144$$I would continue with arithmetic, not logarithms. Girlfriend know$$4^5 = 2^10 = 1,024$$so$$4^10 = 2^20 approx 1,000,000$$is practically four times also big. So $4^9 = 262,144$ and also you need $9$ terms.

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answered Feb 14 at 0:37
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Ethan BolkerEthan Bolker
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That is correct and also it is a reasonable method to fix the problem. Of course you may include up the terms until you get to the answer as result of fast expansion of the terms.


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answer Feb 14 at 0:41
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Mohammad Riazi-KermaniMohammad Riazi-Kermani
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