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Mathematics 18 Online
OpenStudy (anonymous):

Find S12 for the series 1 + 2 + 4 + 8 +...

Parth (parthkohli):

\(\Large \color{Black}{\Rightarrow 2^{n -1} }\) Is the formula. \(\Large \color{Black}{\Rightarrow 2^{12 - 1} }\) In this case.

Parth (parthkohli):

What is 2^11. 1024 * 2 2028

Parth (parthkohli):

Do we have to find the sum or what?

OpenStudy (anonymous):

mo just the S12 of the sequence

Parth (parthkohli):

Then you have it :P

OpenStudy (anonymous):

2028?

Parth (parthkohli):

2048*

OpenStudy (anonymous):

but why did you put 2^11 when it should be S12?

OpenStudy (anonymous):

:S

Parth (parthkohli):

Because S1 is 2^0 S2 is 2^1 S3 is 2^2 So it's one less power.

Parth (parthkohli):

I think it's the sum

OpenStudy (anonymous):

okay

Parth (parthkohli):

I see the plus signs. The sum would be: \(\Large \color{Black}{\Rightarrow 1 \times {1 - 2^{12} \over 1 - 2 } }\)

Parth (parthkohli):

2 is r 12 is n.

Parth (parthkohli):

1 is a(the first term)

OpenStudy (cwtan):

The general formula is \[\frac {a(r^n -1)}{r-1}\] for r>1,r<-1 \[\frac{a(1-r^n)}{1-r}\] for -1<r<1

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