For every integer m from 1 to 100, inclusive, the mth term

Question

For every integer m from 1 to 100, inclusive, the m th  term of a certain sequence is given by (-1)^m*2^(-m). If N is the sum of the first 100 terms in the sequence, then N is

(A) less than -1
(B) between -1 and -1/2
(C) between -1/2 and 0
(D) between 0 and 1/2
(E) greater than 1/2

Any idea how solve this question? I have got no clue how to.

Bunuel · Accepted Answer

1st term  (-1)^1*2^{-1}=-\frac{1}{2} ;
2nd term  (-1)^2*2^{-2}=\frac{1}{4} ;
3rd term  (-1)^3*2^{-3}=-\frac{1}{8} ;
4th term  (-1)^4*2^{-4}=\frac{1}{16} ;
5th term  (-1)^5*2^{-5}=-\frac{1}{32} ;
6th term  (-1)^6*2^{-6}=\frac{1}{64} ;
...

The sum of 1st and 2nd =  -\frac{1}{4} ;
The sum of 3rd and 4th =  -\frac{1}{16} ;
The sum of 5th and 6th =  -\frac{1}{64} ;
...

So, we should basically sum 100/2=50 terms which form a geometric progression with the first term of  -\frac{1}{4}  and common ratio of  \frac{1}{4} :  -\frac{1}{4} ,  -\frac{1}{16} ,  -\frac{1}{64} , ...

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

So, in our case, since the # of terms is large enough, the sum will be very close to  sum=\frac{-\frac{1}{4}}{1-\frac{1}{4}}=-\frac{1}{3} .

Answer: C.

Stiv · Answer

I don't get it. Isn't the first term -1/2 and the common ratio 1/2?
And why do you need to sum up 1 & 2, 3 & 4, 5 & 6 term... etc.?

slay3r · Answer

Well both the solution would fetch you the same result of -1/3 if you consider it as an infinite series.. 
and that would solve the question.

KarishmaB · Answer

Check out posts on this concept on the blog link given in my signature below.

Ousmane · Answer

Why should we consider this as an infinite sequence?

Bunuel · Answer

We are not saying that it's an infinite sequence. But if it were, then the sum would be -1/3 and since we have the sum of the large enough number of terms (100), then the actual sum would be very close to -1/3.

gmacforjyoab · Answer

Lets say we have only the first term m=1 , then the sum of the series (since it is just one term) , would be -1/2
Now , as explained by Bunuel , lets suppose there are infinite terms , in which case the sum is -1/3 . Hence sum of 1 to 100 terms SHOULD lie between these two range "-1/2 and -1/3".

-Jyothi

GMATtracted · Answer

A general term of the expression is:-
 (-1)^m.2^{-m} 
Hence, the 1st 100 terms can be written as:-
 t_1=\frac{-1}{2} 
 t_2=\frac{1}{4} 
 t_3=\frac{-1}{8} 
.
.
.
.
.
 t_{100}=\frac{1}{2^{100}} 
Now, we know that the sum of the first "n" terms of a Geometric progression with a common ratio r<1 is:
 \frac{a(1-r^n)}{1-r} 
where a=1st term, r=common ratio and n=no. of terms.
Now for this case the expression would be(after putting all the values):-
 N=\frac{\frac{-1}{2} }{1-(\frac{-1}{2})} 
 =\frac{\frac{-1}{2} }{\frac{3}{2}} 
 =\frac{-1}{3}.  
Now, since  2^{100}  is a very large number compared to 1,
Hence, we can cay that  1-(\frac{1}{2^{100}})  is slightly less than 1
Hence, N is slightly less than  \frac{-1}{3} .

radhikakhemka007 · Answer

Bunuel  Why don't we use the formula a(r^n-1)/r-1?

rahul16singh28 · Answer

We can use this formula but the result will be same i.e. -1/3

fireagablast · Answer

kind of a janky problem, but the best way to solve is to just start to plug in numbers and see what the pattern is
m=1 --> (-1)^1 * 2(-1) = -1/2
m=2 --> (-1)^2 * 2(-2) = 1/4
m=3 --> (-1)^3 * 2(-3) = -1/8
..
m=100 --> (-1)^100 * 2(-100) =1/(2^100)

your largest number is going to be -1/2, and every EVEN value of m brings the n closer to 0, and every odd number brings it back closer to -1/2. so -1/2<=x<0
m=2;n sum = -1/2+1/4 = -1/4
m=3; n sum = -1/2+1/4-1/8 = -1/4-1/8 = -3/8
etc

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For every integer m from 1 to 100, inclusive, the mth term

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For every integer m from 1 to 100, inclusive, the mth term

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