Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 01 Aug 2009
Posts: 19

For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
07 Jan 2010, 06:22
Question Stats:
62% (02:20) correct 38% (02:36) wrong based on 1512 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




Math Expert
Joined: 02 Sep 2009
Posts: 59180

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
07 Jan 2010, 07:30
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 isA. 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 First of all we see that there is set of 10 numbers and every even term is negative. Second it's not hard to get this numbers: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... enough for calculations, we see pattern now. And now the main part: adding them up is quite a job, after calculations you'll get \(\frac{341}{1024}\). You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than \(\frac{1}{4}\) and less than \(\frac{1}{2}\), so answer is D. 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 \(sum=\frac{b}{1r}\), 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. Answer: D. Other solutions at: sequencecananyonehelpwiththisquestion88628.html#p668661
_________________




Manager
Joined: 27 Apr 2008
Posts: 158

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
07 Jan 2010, 09:13
Alternative, if you use the geometric series formula.
S = \(\frac{a(1r^n)}{1r}\)
where a = first term, r = multiple factor, n = # of terms.
So in this case, a = 1/2, r = 1/2, n = 10.
Therefore,
S=( (1/2) * (1 (1/2)^10) ) / (1(1/2) S=( (1/2) * (1  (1/1024) ) / (3/2) S=(11/1024) / 3 S=(1023/1024 ) / 3 Since 1023/1024 is close to 1, dividing it by 3 would get us to approximately 1/3, which is between 1/2 and 1/4. So the answer is D.




Manager
Joined: 04 May 2010
Posts: 74
WE 1: 2 yrs  Oilfield Service

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
10 Jul 2010, 01:10
Kth term: A(k) = (1) ^ (k+1) * 1 / 2^k
T is the sum of the first 10 terms of this sequence.
A(1) = (1)^(1+1)*1/2^1 = 1/2
A(2) = (1)^(1+2)*1/2^2 = 1/4
You can see that for odd k it would be positive values of \(1/2^k\) and for even k it would be negative values of \(1/2^k\)
So it's like this:
\(T = \frac{1}{2}  \frac{1}{4} + \frac{1}{18}  \frac{1}{16}+ ...\)
Now look for a pattern in the sums of various number of terms.
i.e. Sum of first 2 terms = 1/2  1/4 = 1/4 Sum of first 3 terms = 1/4 + 1/8 = 3/8 Sum of first 4 terms = 3/8  1/16 = 5/16 Sum of first 5 terms = 5/16 + 1/32 = 11/32 and so on...
You can see that all these sums are between 1/4 and 1/2.
On the GMAT when you have tested such a question for about half the terms and you have established a pattern you can safely assume it will continue. This is because this series is analogous to how the GMAT itself adapts the difficulty of its questions based on your answers. It sort of zigzags in a diminishing pattern. So as you increase k, you will be varying only slightly around a sort of ultimate stagnant value at infinity which you could determine as \(\frac{a}{(1  r)}\) i.e. the sum of an infinite Geometric Progression with first term a and common ratio r.
Here we have first term 1/2 and common ratio 1/2
=> Infinite sum = \(\frac{1}{2*(1  (1/2))}= \frac{1}{2*1.5} = \frac{1}{3}\)
With a large number of terms such as 10 terms, your series sum would actual TEND to stagnate around this infinite sum.
Pick D



Manager
Joined: 30 May 2010
Posts: 161

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
18 Aug 2010, 00:17
Initially, I thought about calculating every term. This is almost always the wrong approach. It is more of a number properties problem than sequences.
For the first term, it alternates between positive and negative. For even k, it is positive 1 * \(1/2^k\) and negative 1 for odd k.
The first term is \(1 * 1/2 = 1/2\)
The second term is \(1 * 1/4 = 1/4\)
\(1/2  1/4 = 1/4\)
The third term is \(1 * 1/8 = 1/8\)
\(1/4 + 1/8 = 3/8\)
Looking at the answer choices, you don't need to continue. Since the denominator is increasing exponentially, the terms added and subtracted are becoming closer to 0. From the first term, we know we will never go above \(1/2\). After subtracting the second term, we know we will never go below \(1/4\).



Manager
Joined: 06 Aug 2010
Posts: 149
Location: Boston

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
01 Nov 2010, 11:41
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.
I have no clue what info has been given and how to use it to derive T.
Kindly post a detailed explanation.
Thanks. Prath. When doing sequence problems, it usually helps to look at at least the first few terms. So in this case: \(a_k = (1)^{k+1} * \frac{1}{2^k}\) This gives us: \(a_1 = \frac{1}{2}\) \(a_2 = \frac{1}{4}\) \(a_3 = \frac{1}{8}\) \(a_4 = \frac{1}{16}\) We can stop there. The first, and largest, term is 1/2, and we then subtract 1/4. We will then add and subtract fractions that will continue to get smaller and smaller. So we can immediately eliminate A, B, and C, since the sum cannot possibly be greater than 1/2. Now we're left with D and E. Note that the sum of the first two terms is 1/4. Then you add another 1/8 and subtract 1/16. This pattern will continue all the way through the tenth term, and you should be able to see that there's no way this sum will become less than 1/4. So the answer is D.



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1827

Re: Sequence: can anyone help with this question
[#permalink]
Show Tags
18 Feb 2011, 13:09
A few things here: * I've never seen a real GMAT question that requires one to know any geometric sequence formulas. Of course there are questions where you might use such formulas, but there will always be a different approach available; * In any sequence question which gives an expression for each term, you'll always want to write down the first few terms using the given expression to work out what the sequence looks like. Here we have: \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{64}, \frac{1}{128}, \frac{1}{256}, \frac{1}{512}, \frac{1}{1024}\) * Now, adding all of these fractions together would take a long time to do. The GMAT *never* requires you to perform any crazy calculations, so there must be a different way to answer the question. Notice the answer choices are only estimates, so we only need to estimate the sum of the first 10 terms. When we want to estimate the value of a sum, we ignore terms that make only a tiny contribution to the sum. The last few terms in our sequence are minuscule compared to the first few, so to get a good estimate, we can completely ignore them; adding, say, the first four terms will give a perfectly good approximation of the sum here (you get 5/16, which is enough to choose the right answer). * The sequence in this question is what is known as an 'alternating sequence'  that is, the terms alternate between positive and negative values. When adding an alternating sequence, you most often want to add your terms in pairs first, grouping one positive and one negative (add the 1st and 2nd term, the 3rd and 4th, and so on). One doesn't need to do this for this question, but it does make the answer a bit easier to see  we'd find our sum is \(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \frac{1}{1024}\) from which you can instantly see the sum is greater than 1/4. Since every term here is tiny after the first two, the sum is certainly less than 1/2.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9795
Location: Pune, India

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
26 Feb 2011, 21:46
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.
I have no clue what info has been given and how to use it to derive T.
Kindly post a detailed explanation.
Thanks. Prath. Using some keen observation, you can quickly arrive at the answer... Terms will be: \(\frac{1}{2}  \frac{1}{4} + \frac{1}{8}  \frac{1}{16} + \frac{1}{32}  ...  \frac{1}{1024}\) For every pair of values: \(\frac{1}{2}  \frac{1}{4} = \frac{1}{4}\) \(\frac{1}{8}  \frac{1}{16} = \frac{1}{16}\) etc... So this series is actually just \(\frac{1}{4} + \frac{1}{16} + ... + \frac{1}{1024}\) So the sum is definitely greater than 1/4. When you add an infinite GP with 1/16 as first term and 1/4 as common ratio, the sum will be \(\frac{\frac{1}{16}}{1\frac{1}{4}} = 1/12\). Here, the sum of terms 1/16 + 1/64 + ... 1/1024 is definitely less than 1/12. Hence the sum is definitely less than 1/2. Answer is (D).
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Status: pursuing a dream
Joined: 02 Jun 2011
Posts: 39
Schools: MIT Sloan (LGO)

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
13 Aug 2011, 12:42
Hello! I was just doing this problem and though I would add some "graphic" approach in case it is useful for someone as it has been for me. You can easily draw a graph of the first points on any series to see if they follow a regular pattern. This one specifically is extremely easy to catch as you draw 3+ points, something like this:
Attachments
GP.jpg [ 18.77 KiB  Viewed 110498 times ]



Manager
Status: On...
Joined: 16 Jan 2011
Posts: 154

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
19 Aug 2011, 16:31
I liked Graphical approach is a good one. BUT  felt there is no need of any calculations T = +(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)Subtract the top and bottom T = 1/4 + 1/2^4 + 1/2^6 + 1/2^8 + 1/2^10Now I know T is greater than 1/4 , but if sum of the remaining is smaller than 1/4, then it will definitely be between 1/4 and 1/2 T = 1/4 + 1/4(1/2^2 + 1/2^4 + 1/2^6 + 1/2^8)So sum of the remaining is smaller than 1/4, so ans. is between 1/4 and 1/2 OA D
_________________
Labor cost for typing this post >= Labor cost for pushing the Kudos Button http://gmatclub.com/forum/kudoswhataretheyandwhywehavethem94812.html



Manager
Joined: 30 Aug 2009
Posts: 217
Location: India
Concentration: General Management

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
12 Jan 2012, 04:35
enigma123 wrote: For any integer k from 1 to 10, inclusive, the kth of a certain sequence is given by [(1)^(k+1)] × (1 / 2^k). If T is the sum of the first 10 terms of 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 ½ E. less than 1/4
Can someone please let me know what is the concept behind this and how to solve? Every odd value of k will make (k+1) even and [(1)^(k+1)] will be +ve so we will have k = 1, [(1)^(1+1)] × (1 / 2^1) = 1*(1/2) = 1/2 k = 2, [(1)^(2+1)] × (1 / 2^2) = 1*1/4 = 1/4 k = 3, [(1)^(3+1)] × (1 / 2^3) = 1 * 1/8 = 1/8 so we can observe we will get alternate +ve and ve value for (1/2)^k (1/2)  (1/4) + (1/8)  (1/16) + (1/32)  (1/64) + (1/128)  (1/256) + (1/512)  (1/1024) [(1/2)+ (1/8) + (1/32)+ (1/128) + (1/512)]  [(1/4)+(1/16)+(1/64)+(1/256)+(1/1024)] = approx 0.3 so D



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9795
Location: Pune, India

Re: For every integer k from 1 to 10, inclusive, the kth term of a certain
[#permalink]
Show Tags
01 Feb 2012, 23:11
gpkk 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 1) greater than 2 2)between 1 & 2 3) between 0.5 and 1 4)between 0.25 and 0.5 5)less than 0.25
Could someone please provide a solution to this problem ? To get a hang of what the question is asking, put values for k right away. Say, k = 1, k = 2 etc You get terms such as (1/2) when k = 1, (1/4) when k = 2 etc T = 1/2  1/4 + 1/8  1/16 +.... + 1/512  1/1024 (Sum of first 10 terms) Of course GMAT doesn't expect us to calculate but figure out the answer using some shrewdness. We have 10 terms. If we couple them up, two terms each, we get 5 groups: T = (1/2  1/4) + (1/8  1/16) ...+ (1/512  1/1024) Tell me, can we say that each group is positive? From a larger number, you are subtracting a smaller number in each bracket. e.g. 1/2 is larger than 1/4 so 1/2  1/4 = 1/4 i.e. a positive number 1/8  1/16 = 1/16, again a positive number. We will get something similar to this: T = 1/4 + 1/16 +.... (all positives) Definitely this sum, T, is greater than 1/4 i.e. 0.25 Now, let's group them in another way. T = 1/2 + ( 1/4 + 1/8) + ( 1/16 + 1/32) ...  1/1024 You will be able to make 4 groups since you left the first term out. The last term will also be left out. Each bracket will give you a negative term 1/4 + 1/8 = 1/8 etc Since the first term is 1/2 i.e. 0.5, we can say that the sum T will be less than 0.5 since all the other terms are negative. So the sum, T, must be more than 0.25 but less than 0.5 Answer (D)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 16 Feb 2012
Posts: 144
Concentration: Finance, Economics

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
26 Apr 2012, 07:55
Bunuel 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 isA. 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 First of all we see that there is set of 10 numbers and every even term is negative. Second it's not hard to get this numbers: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... enough for calculations, we see pattern now. And now the main part: adding them up is quite a job, after calculations you'll get \(\frac{341}{1024}\). You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than \(\frac{1}{4}\) and less than \(\frac{1}{2}\), so answer is D. 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[/m], is \(sum=\frac{b}{1r}\), 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. Other solutions at: sequencecananyonehelpwiththisquestion88628.html#p668661 \(\frac{first \ term}{1constant}\) Is this formula reversed when we have an increase by 0<constant<1? Does it look like this \(\frac{first \ term}{1+constant}\)?
_________________
Kudos if you like the post!
Failing to plan is planning to fail.



Math Expert
Joined: 02 Sep 2009
Posts: 59180

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
26 Apr 2012, 08:07
Stiv wrote: \(\frac{first \ term}{1constant}\) Is this formula reversed when we have an increase by 0<constant<1? Does it look like this \(\frac{first \ term}{1+constant}\)? The sum of infinite geometric progression with common ratio \(r<1\), is \(sum=\frac{b}{1r}\), where \(b\) is the first term.
_________________



Director
Joined: 09 Jun 2010
Posts: 697

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
17 Nov 2012, 03:05
gmat dose not test the memory of formular.
HOw to solve in 2 minutes. ?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9795
Location: Pune, India

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
18 Nov 2012, 11:15
thangvietnam wrote: gmat dose not test the memory of formular.
HOw to solve in 2 minutes. ? Check out the explanation of this question here: http://www.veritasprep.com/blog/2012/03 ... sequences/
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Director
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 990
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

Re: For every integer k from 1 to 10, inclusive, the kth term of a certain
[#permalink]
Show Tags
15 Jan 2013, 07:16
an alternative: The series can be written as \(1/2  1/4 + 1/8  1/16.............1/1024\) Since the terms beyond \(1/16\) are too small, so I am not considering those terms. Now on adding \(1/2  1/4 +1/8  1/16\), we get \(5/16\) which is slightly more than \(4/16\). Hence its value would be around `\(0.3\). The answer choice which includes this number is D.
_________________



VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1013
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
23 Apr 2013, 00:51
kabilank87 wrote: For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?
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
I answer this question in my second attempt, but it takes a lot of times around 4 minutes. Please explain a shorter way to do this ? Here we need to find a pattern \(\frac{1}{2},\frac{1}{4},\frac{1}{8},...\) as you see the sign changes every term. The first and bigger is 0,5 and then we subtract and sum smaller and smaller terms. We can eliminate any option that gives us a upper limit greater than 1/2. We are down to D and E. Is the sum less than 1/4? Take the sum of pair of terms : the first 2 give us \(\frac{1}{4}\), the second pair is \(\frac{1}{8}\frac{1}{16}\) positive so we add value to \(\frac{1}{4}\), so the sum will be greater.(this is true also for the next pairs, so we add to \(\frac{1}{4}\) a positive value for each pair) D Hope its clear, let me know
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture  Review: MGMAT workshop Strategy: SmartGMAT v1.0  Questions: Verbal challenge SC III CR New SC set out !! , My QuantRules for Posting in the Verbal Forum  Rules for Posting in the Quant Forum[/size][/color][/b]



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1827

For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
Updated on: 07 Oct 2019, 01:10
kabilank87 wrote: For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?
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
You could learn a formula for such questions (this is an alternating geometric series), but you don't need to. It's almost always a good idea to write out at least the first few terms of a sequence often you'll notice a pattern that will help to answer the question. If you do this, you'll see that the question is asking: 1/2  1/4 + 1/8 1/16 + ... + 1/(2^9)  1/2^(10) = ? There are many ways to estimate the value of this sum. You can notice, for example, that 1/2  1/4 = 1/4; 1/8  1/16 = 1/16, and so on, so the sum is exactly equal to: 1/4 + 1/16 + 1/64 + 1/256 + 1/1028 and which is clearly only slightly greater than 1/4. There's another way you could look at this, though it becomes much more clear when you can draw everything on a number line, which I can't do here. Label the points 0 and 1 on a number line. Suppose you're going to do the following: run halfway from 0 to 1, arriving at A: then A is 1/2 run backwards half the distance between A and 0, arriving at B: then B is 1/2  1/4 run forwards half the distance between B and A, arriving at C: then C is 1/2  1/4 + 1/8 run backwards half the distance between C and B, arriving at D: then D is 1/2  1/4 + 1/8  1/16 etc. you'll see that no matter how many terms you add of this sequence, the sum must be between 1/4 and 1/2.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Originally posted by IanStewart on 23 Apr 2013, 02:01.
Last edited by Bunuel on 07 Oct 2019, 01:10, edited 3 times in total.



Manager
Joined: 12 Dec 2012
Posts: 213
Concentration: Leadership, Marketing
GMAT 1: 540 Q36 V28 GMAT 2: 550 Q39 V27 GMAT 3: 620 Q42 V33
GPA: 2.82
WE: Human Resources (Health Care)

Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
Show Tags
24 Apr 2013, 12:21
Zarrolou wrote: kabilank87 wrote: For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?
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
I answer this question in my second attempt, but it takes a lot of times around 4 minutes. Please explain a shorter way to do this ? Here we need to find a pattern \(\frac{1}{2},\frac{1}{4},\frac{1}{8},...\) as you see the sign changes every term. The first and bigger is 0,5 and then we subtract and sum smaller and smaller terms.We can eliminate any option that gives us a upper limit greater than 1/2. We are down to D and E. Is the sum less than 1/4? Take the sum of pair of terms : the first 2 give us \(\frac{1}{4}\), the second pair is \(\frac{1}{8}\frac{1}{16}\) positive so we add value to \(\frac{1}{4}\), so the sum will be greater.(this is true also for the next pairs, so we add to \(\frac{1}{4}\) a positive value for each pair) D Hope its clear, let me know Can you please elaborate more , Zarrou ? I still do not understand , and the line in red tricks me a lot . Thanks in advance
_________________




Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
[#permalink]
24 Apr 2013, 12:21



Go to page
1 2 3
Next
[ 47 posts ]



