geometric series: evaluate sum, n=2 to n=7, of 2.3^n

[mdem1234]

I have

7
sigma 2.3^n
n=2

and i need to evaluate it.

The answer I get is

r = ratio = 2.3
n = number of terms = 6
a= first term = 5.29

so using the formula S = a(r^n - 1/r-1) I get 5.29 (2.3^6 -1/2.3 -1) = 598.32

The answer is totally different to what I get, it says the answer should be 6552... Am I doing something totally wrong or are the answers i was given wrong? pls help

 

[BigGlenntheHeavy]

Re: geometric series

This is small enough that one can do it on his trusty TI-89.

\(\displaystyle \sum_{n=2}^{7} 2.3^{n} = 2.3^{2}+2.3^{3}+2.3^{4}+2.3^{5}+2.3^{6}+2.3^{7} = 598.3229637\)

 

[soroban]

Hello, mdem1234!

I think you copied it wrong . . . That's not a decimal point.

\(\displaystyle \text{Evaluate: }\:\sum^7_{n=2} 2\!\cdot\!3^n\)

\(\displaystyle \text{We have: }\;S \;=\;2\!\cdot\!3^2 + 2\!\cdot\!3^3 + 2\!\cdot\!3^4+2\!\cdot\!3^5 +2\!\cdot\!3^6 + 2\!\cdot\!3^7\)


\(\displaystyle \text{This is a geometric series with: }\:\begin{Bmatrix}a &=& 2\!\cdot\!3^2 \\ r &=& 3 \\ n &=& 6 \end{Bmatrix}\)


\(\displaystyle \text{The sum is: }\:S \;=\;2\!\cdot\!3^2\left(\frac{1-3^6}{1-3}\right) \;=\;18\left(\frac{\text{-}728}{\text{-}2}\right) \;=\;18(364) \;=\;\boxed{6552}\)