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

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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
Hi

Well we all understood that series is $$\frac{1}{2}$$ , $$\frac{-1}{2^2}$$ , $$\frac{1}{2^3}$$, $$\frac{-1}{2^4}$$, $$\frac{1}{2^5}$$ ,$$\frac{-1}{2^6}$$,$$\frac{1}{2^7}$$ ,$$\frac{-1}{2^8}$$ ,$$\frac{1}{2^9}$$ ,$$\frac{-1}{2^10}$$

Can i rewrite the series as

{$$\frac{1}{2}$$ +$$\frac{1}{2^3}$$ +$$\frac{1}{2^5}$$ +$$\frac{1}{2^7}$$ +$$\frac{1}{2^9}$$ } - { $$\frac{1}{2^2}$$ + $$\frac{1}{2^4}$$ + $$\frac{1}{2^6}$$+ $$\frac{1}{2^8}$$ + $$\frac{1}{2^10}$$ }

{$$\frac{1}{2}$$ +$$\frac{1}{2^3}$$ +$$\frac{1}{2^5}$$ +$$\frac{1}{2^7}$$ +$$\frac{1}{2^9}$$ } -$$\frac{1}{2}$$ { $$\frac{1}{2}$$ +$$\frac{1}{2^3}$$ +$$\frac{1}{2^5}$$ +$$\frac{1}{2^7}$$ +$$\frac{1}{2^9}$$ }

Let {$$\frac{1}{2}$$ +$$\frac{1}{2^3}$$ +$$\frac{1}{2^5}$$ +$$\frac{1}{2^7}$$ +$$\frac{1}{2^9}$$ } =q
then
$$q-\frac{q}{2}$$

$$\frac{q}{2}$$

now q is in GP and we need to find the sum of GP whose first term is $$\frac{1}{2}$$ and r is $$\frac{1}{4}$$ , n is 5
Either you can use the formula for sum of infinte terms of GP or genereal formula .

Lets try the general formula which says sum of n terms of GP

we have $$\frac{\frac{1}{2} * (1 - \frac{1}{4^5})}{ (1-\frac{1}{4})}$$

$$\frac{\frac{1}{2} * ( \frac{4^5-1}{4^5})}{ (\frac{3}{4})}$$

Can we say $$\frac{4^5-1}{4^5}$$ is nearly equal to 1

So this expression $$\frac{\frac{1}{2} * ( \frac{4^5-1}{4^5})}{ (\frac{3}{4})}$$ hence simplifies to $$\frac{2}{3}$$

now q= $$\frac{2}{3}$$

we need the value of $$\frac{q}{2}$$ = $$\frac{1}{3}$$

So Choice D suits our Answer.

PS: Does this solution look long - Yeah because i tried to explain each step. I am sure we will not being many steps that i wrote down . I did get my solution in 1: 03 min

abhimahna

I guess you missed on signs in this post , lucky you ended getting the right answer.

https://gmatclub.com/forum/kth-term-of- ... l#p1724824
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
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prathns 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

Series has 10 terms where first term $$= \frac{1}{2}$$ and 10th term $$= - \frac{1}{2^{10}}$$.
Every odd is positive and every even term is negative.

The series becomes
$$\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \frac{1}{32} - \frac{1}{64} + \frac{1}{128} - \frac{1}{256} + \frac{1}{512} - \frac{1}{1024}$$

Separating positive terms and negative terms we have
$$\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \frac{1}{32} - \frac{1}{64} + \frac{1}{128} - \frac{1}{256} + \frac{1}{512} - \frac{1}{1024}$$
$$= \frac{1}{2}(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) - \frac{1}{4}(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..)$$
$$= (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) * (\frac{1}{2} - \frac{1}{4})$$
$$= \frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..)$$

And
$$1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} .. > 1.25$$
$$\frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) > \frac{1}{4} * 1.25$$
$$\frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) > 0.31$$

Checking answers only D is correct since $$\frac{1}{4} < 0.31 < \frac{1}{2}$$

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For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
Bunuel wrote:

BUT there is shortcut:

Sequence $$\frac{1}{2}$$, $$-\frac{1}{4}$$, $$\frac{1}{8}$$, $$-\frac{1}{16}$$, $$\frac{1}{32}$$... represents geometric progression with first term $$\frac{1}{2}$$ and the common ratio of $$-\frac{1}{2}$$.

Now, the sum of infinite geometric progression with common ratio $$|r|<1$$, is $$sum=\frac{b}{1-r}$$, where $$b$$ is the first term.

So, if the sequence were infinite then the sum would be: $$\frac{\frac{1}{2}}{1-(-\frac{1}{2})}=\frac{1}{3}$$

This means that no matter how many number (terms) we have their sum will never be more then $$\frac{1}{3}$$ (A, B and C are out). Also this means that the sum of our sequence is very close to $$\frac{1}{3}$$ and for sure more than $$\frac{1}{4}$$ (E out). So the answer is D.

Bunuel How do we know that the sum of this sequence will be very close to 1/3 ?
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
Bunuel wrote:

BUT there is shortcut:

Sequence $$\frac{1}{2}$$, $$-\frac{1}{4}$$, $$\frac{1}{8}$$, $$-\frac{1}{16}$$, $$\frac{1}{32}$$... represents geometric progression with first term $$\frac{1}{2}$$ and the common ratio of $$-\frac{1}{2}$$.

Now, the sum of infinite geometric progression with common ratio $$|r|<1$$, is $$sum=\frac{b}{1-r}$$, where $$b$$ is the first term.

So, if the sequence were infinite then the sum would be: $$\frac{\frac{1}{2}}{1-(-\frac{1}{2})}=\frac{1}{3}$$

This means that no matter how many number (terms) we have their sum will never be more then $$\frac{1}{3}$$ (A, B and C are out). Also this means that the sum of our sequence is very close to $$\frac{1}{3}$$ and for sure more than $$\frac{1}{4}$$ (E out). So the answer is D.

Bunuel How do we know that the sum of this sequence will be very close to 1/3 ?

Bunuel's explanation included a formula that shows why the sum gets fairly close to 1/3. Looking at the answer choices though, you do NOT actually have to get to that conclusion (since the answers are RANGES, we just need to do enough work to define which "range" the total will be in - and then we can stop working).

By plugging in the first few numbers (1, 2, 3, 4) into the sequence, you can see that a pattern emerges among the terms....

1st term = +1/2
2nd term = -1/4
3rd term = +1/8
4th term = -1/16

The terms follow a positive-negative-positive-negative pattern all the way to the 10th term and each term is the product of the prior term and 1/2. By "pairing up' the terms, another pattern emerges....

1/2 - 1/4 = 1/4

1/8 - 1/16 = 1/16

1/32 - 1/64 = 1/64

Etc.

The "pairs" get progressively smaller (notice how each is the product of the prior term and 1/4). This means that we're "starting with" 1/4 and adding progressively TINIER fractions to it. Since we're just adding 4 fractions (that are getting REALLY SMALL) to 1/4, this means that we're going to end up with a total that's just a LITTLE MORE than 1/4. Looking at the answer choices, there's only one answer that fits:

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For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
prathns wrote:
For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by $$(-1)^{(k+1)}*(\frac{1}{2^k})$$. 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

Given: For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by $$(-1)^{(k+1)}*(\frac{1}{2^k})$$.
Asked: If T is the sum of the first 10 terms in the sequence, then T is

$$T = \frac{1}{2} - \frac{1}{4} + \frac{1}{8 }- \frac{1}{16} + ......- \frac{1}{2^{10}}$$

$$T_{max} = \frac{1}{2} - \frac{1}{4} + ...... to\ infinity = \frac {\frac{1}{2} }{1+ \frac{1}{2}} =\frac{1}{3}$$

IMO D
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
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prathns wrote:
For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by $$(-1)^{(k+1)}*(\frac{1}{2^k})$$. 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

Solution:
We are given that for every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (1/2^k). We must determine the sum of the first 10 terms in the sequence. Before calculating the sum, we should recognize that the answer choices are provided as ranges of values, rather than as exact values. Thus, we might not need to calculate the total of the 10 terms to determine an answer. Perhaps we can uncover a pattern to help us find the answer. Let’s start by listing the first four terms.
k = 1:
(-1)^(1+1) x (1/2^1)
(-1)^2 x 1/2
1 x 1/2 = 1/2
k = 2:
(-1)^(2+1) x (1/2^2)
(-1)^3 x 1/4
-1 x 1/4 = -1/4
k = 3:
(-1)^(3+1) x (1/2^3)
(-1)^4 x 1/8
1 x 1/8 = 1/8
k = 4:
(-1)^(4+1) x (1/2^4)
(-1)^5 x 1/16
-1 x 1/16 = -1/16
Recall that we are trying to estimate the value of T = 1/2 + (-1/4) + 1/8 + (-1/16) + … until we have 10 terms. In other words, T = 1/2 – 1/4 + 1/8 – 1/16 + … until there are 10 terms.
We should notice that the absolute values of the terms are getting smaller:
|1/2|>|-1/4|>|1/8|>|-1/16|.
Notice that starting from the first term of 1/2, we are subtracting something less than 1/2 (notice that 1/4 < 1/2) but then adding back something even less (notice 1/8 < 1/4), and the process continues. Thus, because ½ and -1/4 are the largest term and the smallest term, respectively, in our set, the sum will never fall below ¼ or exceed ½.
Thus, we conclude that T is greater than 1/4 but less than 1/2.

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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]
prathns wrote:
For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by $$(-1)^{(k+1)}*(\frac{1}{2^k})$$. 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

Let us first look at the terms by substituting k = 1, 2, 3, 4 ... :
1/2, -1/4, 1/8, -1/16 ...

Observe that the options are in ranges and so, we need not necessarily calculate the exact value of this summation. Also, both 1/2 and 1/4 are there in the options - the first 2 terms of the series above

Adding the first 2 terms, we get 1/2 - 1/4 = 1/4
Thereafter, the sum of every pair is a positive quantity, for example, 1/8 - 1/16 > 0 etc.
Thus, the sum is greater than 1/4

Again, leaving out the first tern, 1/2, if we pair, each pair would be negative (for example: -1/4 + 1/8 < 0) and the 10th term would be left over, which also is negative.
Thus, the sum is less than 1/2

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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain [#permalink]