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question on sequences

by stephkhaira » Mon Aug 26, 2013 6:38 pm
For all 1<=k<=10, the kth term of a sequence is (-1)^(k+1) * (1/2^k). If T is the sum of the first 10 terms in the sequence, then T is

a) Greater than 2
b) Between 1 and 2
c) Between 0.5 and 1
d) Between .25 and .5
e) Less than .25



[spoiler]ans: d) Between .25 and .5[/spoiler]
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by Brent@GMATPrepNow » Mon Aug 26, 2013 6:49 pm
stephkhaira wrote:For all 1<=k<=10, the kth term of a sequence is (-1)^(k+1) * (1/2^k). If T is the sum of the first 10 terms in the sequence, then T is

a) Greater than 2
b) Between 1 and 2
c) Between 0.5 and 1
d) Between .25 and .5
e) Less than .25

[/spoiler]
One way to solve this question is to apply the formula for what's known as a "geometric series," but I'm not really a fan of memorizing formulas. Another option is to get a better idea of this sum, by first finding a few terms:

T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
We can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)

Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less than 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024)
= 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, [spoiler]1/4 < T < 1/2[/spoiler]

Answer: D

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Brent
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by GMATGuruNY » Mon Aug 26, 2013 7:15 pm
For every integer K from 1 to 10, inclusive the kth term of a certain sequence is given by (-1)^(k+1) (1/2^K).
If T is the sum of first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to 1/2
E. Less than ¼
Calculate until you see the pattern.
Some test-takers might find it helpful to visualize the sum on a number line.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (1/16).
In other words, the sum will alternate between increasing a little and then decreasing a little less than it went up.

The sum of the first 2 terms is 1/4. From there, the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. Here are the first four terms, plotted on a number line:

Image

Since all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is D.
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by sanjoy18 » Fri Aug 30, 2013 9:23 am
1<=K<=10
T1=(-1)^2*1/2=1/2
T2=(-1)^3*1/2^2=-1/2^2
T1+T2+T3+...................+T10
[{(1/2)-(1/2^2)}+{(1/2^3)-(1/2^4)}+....................+{(1/2^9)-(1/2^10)}]
= (1/2^2)+(1/2^4)+(1/2^6)+(1/2^8)+(1/2^10)
USING GP FORMULA
= (1/2^2)(1-(1/2^2)^5)/(1-(1/2^2))
=(1/4)*(2^10-1)*4/(2^10*3)
=(2^10-1)/(2^10*3)
=(1/3)*(1-(1/2^10))
=~ 1/3 = .333 (1/1024 Is negligible w.r.t two decimal place)
=.33
= between .25 and .5
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by Mission2012 » Fri Aug 30, 2013 6:15 pm
sanjoy18 wrote:1<=K<=10
T1=(-1)^2*1/2=1/2
T2=(-1)^3*1/2^2=-1/2^2
T1+T2+T3+...................+T10
[{(1/2)-(1/2^2)}+{(1/2^3)-(1/2^4)}+....................+{(1/2^9)-(1/2^10)}]
= (1/2^2)+(1/2^4)+(1/2^6)+(1/2^8)+(1/2^10)
USING GP FORMULA
= (1/2^2)(1-(1/2^2)^5)/(1-(1/2^2))
=(1/4)*(2^10-1)*4/(2^10*3)
=(2^10-1)/(2^10*3)
=(1/3)*(1-(1/2^10))
=~ 1/3 = .333 (1/1024 Is negligible w.r.t two decimal place)
=.33
= between .25 and .5
Hi Sanjoy,

if T1 = 1/2, T2 = -1/4, T3 = 1/8, T4 = -1/16....

Is it not a gp with a multiplier of -1/2.

Using formula for GP

T = 1/2(1- (-1/2)^10)/(1-(-1/2))
= 1/2(1- (-1/2)^10)/(3/2))
= 1/3(1 - (-1/2)10)
= ~1/3

Isnt this approach also correct
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by sanjoy18 » Sat Aug 31, 2013 2:51 am
Hi Mission 2012

yeah ..your approach is also correct..

Basically..In finite GP series..there is two formula

1) Sum= a(1-r^n)/(1-r) where |r|<1
2) sum= a(r^n-1)/(r-1) where |r|>1
for your case it first one..where r=-1/2
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