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

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Math Expert
Joined: 02 Sep 2009
Posts: 56307
For every integer n from 1 to 200, inclusive, the nth term of a certai  [#permalink]

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02 Apr 2018, 02:28
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Difficulty:

55% (hard)

Question Stats:

66% (02:14) correct 34% (02:19) wrong based on 73 sessions

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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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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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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

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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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.

*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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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.

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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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06 Apr 2019, 09:31
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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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