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=

Answer 1

The given sequence is a geometric sequence with a common ratio of (1/3). So, the (n)th term of the sequence can be expressed as (a_n = (1/3)^n), where (n) represents the position of the term in the sequence. Therefore, the sequence continues indefinitely as (1/3^n) for (n = 1, 2, 3, \ldots).

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Answer 2

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

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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