How do you find the sum of the geometric series #Sigma 1/4*2^(n-1)# n=1 to 5?

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Answer 1 · 2016-12-16T06:14:02

Sum of first five terms #n=1# to #n=5# of given series is #7 3/4#

In the geometric series #Sigma 1/4*2^(n-1)#

first term #a_1# is #a_1=1/4xx2^(1-1)=1/4xx1=1/4#

Subsequent terms are #a_2=1/4xx2^(2-1)=1/4xx2=1/2#

and #a_3=1/4xx2^(3-1)=1/4xx4=1#

Observe that the power of #2# increases by one from a term to the next term,

Hence common ration #r=2#

As sum of a geometric series #S_n#, with first term as #a_1# and common ratio as #r# is

#S_n=(a_1(r^n-1))/(r-1)#

Sum of first five terms #n=1# to #n=5# of given series is

#S_5=(1/4(2^5-1))/(2-1)=1/4xx(32-1)/1=1/4xx31=31/4=7 3/4#