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Mathematics 27 Online
OpenStudy (driftracer305):

what is the sum of the geometric series 15 E x=0 2(1/2)^x........... this is summation notation here..!!

OpenStudy (driftracer305):

@ganeshie8

OpenStudy (driftracer305):

@jim_thompson5910

OpenStudy (driftracer305):

@Luigi0210

OpenStudy (driftracer305):

@mathstudent55 ............sum help would be great....

OpenStudy (campbell_st):

well find the first few terms x = 0 2 x = 1 1 x = 2 1/2 x = 3 1/4 so the 1st term is 2, the common ratio is r = 1/2 and the number of terms is n = 16 since you are going from 0 to 15 so use the formula \[S_{n} = \frac{a(1 - r^n)}{(1 - r)}\] hope it helps

OpenStudy (dls):

right^

OpenStudy (driftracer305):

@campbell_st ........thanks a TON.......... just had some doubts.....

OpenStudy (driftracer305):

k thx.... appreciate it

OpenStudy (dls):

\[\Huge \sum_{0}^{15} \frac{2}{2^x}= 2+ 1 + \frac{1}{2} + \frac{1}{4}....\] 3+ (sum of a GP)

OpenStudy (dls):

\[\Huge S_n = \frac{a(1- r^n)}{1-r}\] where a is your first term and r the common difference

OpenStudy (driftracer305):

should be 4 right...?

OpenStudy (dls):

\[\LARGE 3+\frac{\frac{1}{2} ( 1- \frac{1}{2^{14}})}{\frac{1}{2}}\] this should be your answer I guess

OpenStudy (driftracer305):

yup... it;s 4......... thanks a lot @DLS

OpenStudy (dls):

yeah one..because \[\LARGE \frac{1}{2^{14}} ->0\] since its a very large quantity :)

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