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# Sum of x +1/x + x^2 + 1/x^2 + x^3 + 1/x^3 + infinity

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Manager
Joined: 11 Dec 2013
Posts: 114
Location: India
GMAT Date: 03-15-2015
WE: Education (Education)
Sum of x +1/x + x^2 + 1/x^2 + x^3 + 1/x^3 + infinity  [#permalink]

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Updated on: 03 Feb 2019, 00:56
1
00:00

Difficulty:

65% (hard)

Question Stats:

29% (01:35) correct 71% (02:01) wrong based on 18 sessions

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Sum of $$(x+\frac{1}{x})+(x^{2}+\frac{1}{x^{2}})+(x^{3}+\frac{1}{x^{3}})+.....\infty$$ is

A) $$\frac{x+1}{-x}$$

B) $$\frac{x}{1+x}$$

C) $$\frac{x}{1-x}$$

D) $$0$$

E) $$-1$$

_________________

Originally posted by 4d on 02 Feb 2019, 19:16.
Last edited by Bunuel on 03 Feb 2019, 00:56, edited 1 time in total.
Renamed the topic and edited the question.
Director
Joined: 18 Jul 2018
Posts: 680
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: Sum of x +1/x + x^2 + 1/x^2 + x^3 + 1/x^3 + infinity  [#permalink]

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02 Feb 2019, 19:44
1
1
The question can be rearranged as $$x+x^2+x^3+$$...... and $$\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+$$......

The above expression is a GP series.
a = x
r = x
Sum of infinite series = $$\frac{a}{1-r}$$ = $$\frac{x}{1-x}$$

For the second expression, a = 1/x, r = 1/x
Sum of infinite series = $$\frac{1}{x/(1-1/x)}$$ = $$\frac{1}{x-1}$$

Adding both, $$\frac{x}{1-x} + \frac{1}{x-1}$$ = $$\frac{x}{1-x} - \frac{1}{1-x}$$ =$$\frac{x-1}{1-x}$$ = -1.

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Press +1 Kudo If my post helps!

Math Expert
Joined: 02 Aug 2009
Posts: 7335
Re: Sum of x +1/x + x^2 + 1/x^2 + x^3 + 1/x^3 + infinity  [#permalink]

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02 Feb 2019, 19:50
4d wrote:
Sum of $$(x+\frac{1}{x})+(x^{2}+\frac{1}{x^{2}})+(x^{3}+\frac{1}{x^{3}})+.....\infty$$ is
A) $$\frac{x+1}{-x}$$
B) $$\frac{x}{1+x}$$
C) $$\frac{x}{1-x}$$
D) $$0$$
E) $$-1$$

Let us convert the equation in a bit of friendly terms...

$$(x+\frac{1}{x})+(x^{2}+\frac{1}{x^{2}})+(x^{3}+\frac{1}{x^{3}})+.....\infty=(x+x^2+x^3+........)+(\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}}+.....)$$
So we have an infinite geometric progression and sum of such progressions is $$\frac{a}{1-r}$$, where a is the first term and r is the ratio by which each term differes with the previos one.
$$=\frac{x}{1-x}+(1/x)/(1-(1/x))=\frac{x}{1-x}+\frac{1}{x-1}=\frac{x}{1-x}+\frac{1}{-(x-1)}=\frac{-(1-x)}{1-x}=-1$$

E
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html

GMAT Expert

Re: Sum of x +1/x + x^2 + 1/x^2 + x^3 + 1/x^3 + infinity   [#permalink] 02 Feb 2019, 19:50
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