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Find S12 for the series 1 + 2 + 4 + 8 +...
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\(\Large \color{Black}{\Rightarrow 2^{n -1} }\) Is the formula. \(\Large \color{Black}{\Rightarrow 2^{12 - 1} }\) In this case.
What is 2^11. 1024 * 2 2028
Do we have to find the sum or what?
mo just the S12 of the sequence
Then you have it :P
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2028?
2048*
but why did you put 2^11 when it should be S12?
:S
Because S1 is 2^0 S2 is 2^1 S3 is 2^2 So it's one less power.
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I think it's the sum
okay
I see the plus signs. The sum would be: \(\Large \color{Black}{\Rightarrow 1 \times {1 - 2^{12} \over 1 - 2 } }\)
2 is r 12 is n.
1 is a(the first term)
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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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