Problem Solving

Hi All,

Could anyone please suggest how to do this in under 2 minutes.?

Thanks,
Mallika

mallika hunsur wrote:


Hi All,

Could anyone please suggest how to do this in under 2 minutes.?

Thanks,
Mallika

[spoiler]NB: Posting the screenshot from the source without consent may put one in legal trouble*[/spoiler]

The expression (-1)^(k + 1) (1/2^k) for integer k, 1 ≤ k ≤ 10 would generate positive and negative terms alternately, resulting in the following shape:

½ - ½^2 + ½^3 - ½^4 + ½^5 - ½^6 + ½^7 - ½^8 + ½^9 - ½^10

= (½ + ½^3 + ½^5 + ½^7 + ½^9) - (½^2 + ½^4 + ½^6 + ½^8 + ½^10)

= slightly more than ½ - slightly more than ¼, which is somewhere between [spoiler]¼ and ½.

Choose (D)[/spoiler]

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 to1/2
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:



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.