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=aigg(fracr^n-1r-1igg)\$\$

Therefore

eginalign*S_n&=2igg(frac4^n-14-1igg)\<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
Laufen 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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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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