Consider the series 4 + 12 + 36 ... Find the sum of the first 9 terms.

Mathematics

1 answer:

jeyben [28]2 years ago

3 0

Answer:

39364

Step-by-step explanation:

You can simplify the equation to make 4(1+3+9...) The nth term is therefore $3^{n-1}$ so u can substitute that into the equation so 4( $3^{n-1}+ 3^{n} + 3^{n+1}$ ..). This would simplify to 4( $3^0 +3+3^2$ + $3^3+3^4+3^5+3^6+3^7+3^8$ ). By simplifying this, you can make it so 4( $2^2+6^2+18^2+42^2+81^2$ ) = 4(4+36+324+2916+6561) =39364

You might be interested in

The temperature is 8° Fahrenheit. If the weatherman predicts the

Zina [86]

Answer:

-4 degrees Farenheit

Step-by-step explanation:

8-12=-4

6 0

3 years ago

Read 2 more answers

You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%

bezimeni [28]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}$

$\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}$