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# For every integer k from 1 to 10, inclusive, the kth term of

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For every integer k from 1 to 10, inclusive, the kth term of [#permalink]

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27 Mar 2006, 13:56
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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 the first 10 terms in the sequence, then T is

A. Greater than 2

B. Between 1 and 2

C. Between 1/2 and 1

D. Between 1/4 and 1/2

E. Less than 1/4

This is a simple one but I'm sure there's a shortcut
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27 Mar 2006, 14:18
a1 = 1/2
a2 = -1/4
a3 = 1/8
a4 = -1/16

So sum of a1..to a10 will be between 1/4 and 1/2
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08 Apr 2006, 22:08
macca wrote:
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 the first 10 terms in the sequence, then T is

A. Greater than 2

B. Between 1 and 2

C. Between 1/2 and 1

D. Between 1/4 and 1/2

E. Less than 1/4

This is a simple one but I'm sure there's a shortcut

Using this given sequence i.e, -1^k+1 /2^k

The sum T equals to: the first term-last the last term
so, 1/2+1/2^10 which is between 1/2 and 1

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09 Apr 2006, 22:55
macca wrote:
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 the first 10 terms in the sequence, then T is

A. Greater than 2

B. Between 1 and 2

C. Between 1/2 and 1

D. Between 1/4 and 1/2

E. Less than 1/4

This is a simple one but I'm sure there's a shortcut

Kth Term = (-1)^(k+1) * (1/2^k)
1st term = 1/2
2nd term = -1/4
3rd term = 1/8
.
.
.
.
so on

Sum of First 10 term = 1st term + 2nd term + 3rd term +...........+ 10th
term
= 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 ...........
= {1/2+ 1/2^3+ 1/2^5+ 1/2^7+ 1/2^9} - { 1/2^2
+ 1/2^4 + 1/2^6 +1/2^8 + 1/2^10}

These are two Geometirc Series in first ,I think you all know the some nth term of geometirc series from that we can say that
= 1/2- 1/4

or between 1/4 and 1/2
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25 Apr 2006, 05:01
This is in the form: 1/2 - 1/4 + 1/8 - 1/16..... until -1/1024

This is in Geometric Progression (G.P.)

Sum of terms in G.P. = a((1-r^n)/(1-r)) (wherer < 1)

where a is initial term, r is the ratio, n is number of terms

here,
a=1/2, r=-1/2, n=10

we will get 1023/(1024 *3)

i.e. greater than 1/4 and lesser than 1/2
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25 Apr 2006, 07:17
klsrikanth wrote:
we will get 1023/(1024 *3)

Well done. That's right. The limit for this series for k -> infinity is 1/3.
25 Apr 2006, 07:17
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