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For every integer n from 1 to 200, inclusive, the nth term of a certai

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For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

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New post 02 Apr 2018, 02:28
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For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by \((-1)^n*2^{(-n)}\). If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

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For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

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New post 02 Apr 2018, 09:44
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Bunuel wrote:
For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by \((-1)^n*2^{(-n)}\). If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2


the sequence can be written as \(\frac{-1}{2}, \frac{1}{2^2}, \frac{-1}{2^3}, \frac{1}{2^4}......................., \frac{-1}{2^{199}}, \frac{1}{2^{200}}\). For simplicity let \(\frac{1}{2}=a\)

so \(N= (a^2+a^4+.....a^{200})-(a+a^3+......a^{199})\). This is a GP series. Sum of GP series is \(a_1*\frac{(1-r^n)}{1-r}\). Here we have two GP series and each has 100 terms, for the first series \(a_1=a^2\) and for the second series \(a_1=a\), common ratio, \(r=a^2\) and \(n=100\), which is same for both the series.

Hence \(N= a^2*\frac{[1-(a^2)^{100}]}{1-a^2} - a*\frac{[1-(a^2)^{100}]}{1-a^2} = \frac{[1-(a^2)^{100}]}{1-a^2}*(a^2-a)\). Now substitute the value of \(a\) here to get

\([(1-\frac{1}{2^{200}})/(1-\frac{1}{2^2})]*(\frac{1}{2^2}-\frac{1}{2})\). on solving this equation, we will get \((\frac{1}{2^{200}}-1)*\frac{1}{3}\)

Among the options given, this value will fall between \(\frac{-1}{2}\) and \(0\)

Option C
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For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

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New post 03 Apr 2018, 00:20
Bunuel wrote:
For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by \((-1)^n*2^{(-n)}\). If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2


SUGGESTION: For any question of sequence, make sure that you write atleast 4-5 terms to understand the pattern in which the sequence is taking shape.

\(T_1 = -1/2\)
\(T_2 = +1/4\)
\(T_3 = -1/8\)
\(T_4 = +1/16\)
\(T_5 = -1/32\) and so on...

i.e. Sum = -1/2+1/4-1/8+1/16-1/32+1/64---- and so on

i.e. Sum = -1/4-1/16-1/64 = -0.25-0.0625-0.015625---

i.e. Sum must be between 0 and -1/2

Answer: Option C

P.S. The formula for sum of a geometric progressions is completely un-necessary for any GMAT question which can be approached using that formula hence reader who haven't seen the formula of sum of a GP need not worry as there is always an alternative that GMAT expects you to know.
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For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

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New post 03 Apr 2018, 09:59
Bunuel wrote:
For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by \((-1)^n*2^{(-n)}\). If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

As GMATinsight suggests, because the geometric approach seemed too time-consuming, I looked for a pattern. This arithmetic is not hard.

The nth term of a certain sequence is given by \((-1)^n*2^{(-n)}\)

First 6 terms
\(A_1 =(-1^1*2^{-1})=(-1*\frac{1}{2^1})=-\frac{1}{2}\)

\(A_2 =(-1^2*2^{-2})=(1*\frac{1}{2^2})=(1*\frac{1}{4})=\frac{1}{4}\)

\(A_3 =(-1^3*2^{-3})=-\frac{1}{8}\)

\(A_4 =(1*2^{-4})=\frac{1}{16}\)

\(A_5 = (-1*\frac{1}{32})=-\frac{1}{32}\)

\(A_6 = (1*\frac{1}{64})=\frac{1}{64}\)


Pattern? Sum a few terms

We need a pattern for the sum of terms.
Extrapolate directly from the terms' values above, or add:

\(A_1+A_2= (-\frac{1}{2}+\frac{1}{4})=-\frac{1}{4}\)

\((A_1+A_2) + A_3= (-\frac{1}{4}+\frac{1}{8})=-\frac{1}{8}\)

\((A_1+A_2+A_3)+ A_4= (-\frac{1}{8}+\frac{1}{16})=-\frac{1}{16}\)


The pattern emerges: Summing the terms gets us closer to 0 by exactly half.

We will never reach 0. We add a positive number to a negative number, but |negative| > positive

\(-Sum + \frac{1}{2}Sum= -\frac{1}{2}Sum\)

The added positive term is never enough to get to 0 and beyond.
Upper limit is 0.
Only one answer has 0 on RHS. That's enough.

Answer C

*Lower limit? Sequence starts with \(A_1 = -\frac{1}{2}\)
But the sum of just 2 terms (let alone 200), \(A_1+A_2\), is already > than \(-\frac{1}{2}\)
Whole sequence, summed, is thus also > than \(-\frac{1}{2}\)
Lower limit is \(-\frac{1}{2}\)
\(-\frac{1}{2} < Sum < 0\)

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Re: For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

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New post 05 Apr 2018, 17:10
Bunuel wrote:
For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by \((-1)^n*2^{(-n)}\). If N is the sum of the first 200 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2



Let’s solve the expression for values of n of 1, 2, 3, and 4.

n = 1:

-1 x 1/2 = -1/2

n = 2:

1 x 1/4 = 1/4

n = 3:

-1 x 1/8 = -1/8

n = 4:

1 x 1/16 = 1/16

Thus, the first 4 terms of the sequence are:

-1/2, 1/4, -1/8, 1/16, …

We can see that this is a geometric sequence with the first term a1 = -1/2 and the common ratio r = -1/2. Recall that the sum of a finite geometric sequence with n terms is S(n) = a1 * (1 - r^n)/(1 - r). Thus, N or S(200), is

-1/2 * (1 - (-1/2)^200)/(1 - (-1/2))

Notice that (-1/2)^200 is very small, so we can approximate it as 0. Thus, we have

-1/2 * (1 - 0)/(1 + 1/2)

(-1/2)/(3/2)

-1/3

Thus we see that N is between -1/2 and 0.

Answer: C
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Re: For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

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Re: For every integer n from 1 to 200, inclusive, the nth term of a certai   [#permalink] 06 Apr 2019, 09:31
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