For the sequence 1/3, 1/3^2 ,1/3^3 ,1/3^4 ,1/3^5,…, ?

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For the sequence 1/3, 1/3^2 ,1/3^3 ,1/3^4 ,1/3^5,…, ?

its fifth partial sum S5=
its sixth partial sum S6=

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Answer 1 · 2016-06-21T15:23:32

#S_5=121/243.#
#S_6=364/729.#

Explanation:

For the Geometric seq. #a,ar, ar^2, ar^3,....#, the sum #S_n# of first #n# terms is given by #S_n={a*(1-r^n))/(1-r).#

Here, #a=1/3, r=1/3,# so, #S_n=[(1/3){1-(1/3)^n}]/(1-1/3),# i.e., #1/2{1-(1/3)^n}.#

Accordingly, #S_5=1/2{1-(1/3)^5}=1/2(1-1/243)=242/(2*243)=121/243.#

For #S_6#, we use a little trick, by observing that, #S_6=S_5+t_6,# [Here, #t_6#=sixth term]
=#121/243+1/3^6=121/243+1/(243*3)=(121*3+1)/(243*3)=364/729#

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For the sequence 1/3, 1/3^2 ,1/3^3 ,1/3^4 ,1/3^5,…, ?

its fifth partial sum S5=
its sixth partial sum S6=

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

For the sequence 1/3, 1/3^2 ,1/3^3 ,1/3^4 ,1/3^5,…, ?

its fifth partial sum S5= its sixth partial sum S6=

1 Answer

Explanation:

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Impact of this question

its fifth partial sum S5=
its sixth partial sum S6=