GMAT PREP QUES?

I am not able to decipher the expression u have said ..
is it
(-1)^k + (1/2)^k

or

(-1)^k + (3/2)^k


Thanks
Senthil

I am not able to decipher the expression u have said ..
is it
(-1)^k + (1/2)^k

or

(-1)^k + (3/2)^k


Thanks
Senthil

Its to the power of k+1

dferm 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 sume of the first 10 terms in the sequence, then T is:



A. greater than 2
B. between 1 and 2
C. between ½ and 1
D. between ¼ and ½
E. less than ¼


Please explain.

Hey guy, check the bolded item above! You said it was ^k+1 so is the equation ((-1)^k+1))(1/(2^k) or ((-1)^k+1)((1/2)^k). It is difficult to HELP YOU if you can't simply post the question correctly.

The sum will be between 1/2 to 1.

This is a geometric progression series.Thereby I feel even if u sum to 20+ numbers it wud not be more than 1.
There for the answer shud be C.

Let me know if I am wrong!

Senthil

The answer is D.

Let me know if I am incorrect.

Does anyone know how to solve this question

I think answer's A, here's why:

formula for sum of numbers in a set is: (n/2)*(first number + last number) (ie 2+3+4+5 = (4/2)*(2+5) = 14.)

So, sum of this sequence would be:

(10/2)*[((-1)^(1+1)*(1/(2^1))] + [((-1)^(1+10)*(1/(2^10))]
=5*[(1)*(1/2)] + (-1)*(1/(2^10))
=5*[(1/2)+(-1/1024)]
=5*(512/1024)-(1/1024)
=5*(511/1024)
=5* something slightly smaller than 1/2, which means T=2.5something, which is greater than 2, thus A.

Any comments/thoughts on this approach?

I just plugged in the first few integers and looked for a pattern. When k=1, the corresponding amount is 1/2, when k=2, it is -1/4, k=3 is 1/8, k=4 is -1/16, and so on. You can stop computing for k here because you can see a pattern.

Now add the ks, and look for another pattern:

I have 1/2 of a t-pizza, you take 1/4 of a pizza, leaving me with 1/4 of a pizza. You give me 1/8 of a pizza back. I have now 1/4 + 1/8 = 3/8. You take 1/16 away, and that leaves me with 5/16.

If you go all the way to k=10, you can never go below 1/4 or above 1/2. Therefore, it's D.

Does this makes sense?

Nevermind, I realized shortly after I wrote my post that my rationale was based on consecutive numbers, and actually has no meaning in this question.

Here's a good explanation by Ron: https://www.manhattangmat.com/forums/for ... t1950.html

Isn't the (-1^1-1)(1/2^1) = -1/2 instead of 1/2
k= 1

brb588 wrote:I just plugged in the first few integers and looked for a pattern. When k=1, the corresponding amount is 1/2, when k=2, it is -1/4, k=3 is 1/8, k=4 is -1/16, and so on. You can stop computing for k here because you can see a pattern.

Now add the ks, and look for another pattern:

I have 1/2 of a t-pizza, you take 1/4 of a pizza, leaving me with 1/4 of a pizza. You give me 1/8 of a pizza back. I have now 1/4 + 1/8 = 3/8. You take 1/16 away, and that leaves me with 5/16.

If you go all the way to k=10, you can never go below 1/4 or above 1/2. Therefore, it's D.

Does this makes sense?

vladmire wrote:Isn't the (-1^1-1)(1/2^1) = -1/2 instead of 1/2
k= 1

You got the equation wrong, it's (-1)^(1+1) * 1/(2^1) =

-1^2 * 1/2 =

1 * 1/2 =

1/2

I ll go for D if u consider (-1)^(K+1) * (1/2)^k..
put K from 1 to 10 and just expand the series...
U'll get something like all the odd powers of 1/2 from 1 to 9 minus all the even powers of 1/2 from 2 to 10. We'll get two GP's with C.R. 1/2. Solving them we'll get answer as 1023/2048 which is less than 1/2 nut greater than 1/4