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Mathematics 7 Online
OpenStudy (anonymous):

Find the 7th partial sum of ...

OpenStudy (anonymous):

\[\sum_{i=1}^{\infty} 6(3) (^i-^1)\]

OpenStudy (anonymous):

the exponent is i-1 I accidentally put an underscore

OpenStudy (anonymous):

can you help me? @xxferrocixx

OpenStudy (anonymous):

I don't know this. :O

OpenStudy (anonymous):

@ganeshie8 @thomaster @Hero

OpenStudy (anonymous):

it's okay, thanks anyways! @jcpd910

OpenStudy (anonymous):

\[ \sum_{i=1}^\infty 6(3)^{i-1} \]

OpenStudy (anonymous):

This is a geometric series.

OpenStudy (anonymous):

When \[a_n = ar^{n-1} \]Then \[ S_n = \frac{1-r^n}{1-r} \]

OpenStudy (anonymous):

Where \(S_n\) is the \(n\)th partial sum: \(S_n = \sum_{i=1}^na_i\)

OpenStudy (anonymous):

\[ \sum_{i=1}^\infty \color{blue}{6}(\color{red}{3})^{i-1} \]\[ a_n =\color{blue}{a}(\color{red}{r})^{n-1} \]

OpenStudy (anonymous):

okay, I get it now. Thank you so much!

OpenStudy (anonymous):

I need to make a correction: When \[a_n = ar^{n-1} \]Then \[ S_n =a\left( \frac{1-r^n}{1-r}\right) \]

OpenStudy (anonymous):

thank you!

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