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Expert replies
by sud21 » Wed Jan 25, 2012 7:26 pm
1/2+1/4+1/8+...+1/512=?
A.0<S<1/2
B.1/2<S<1
C.1<S<3/2
D. 3/2<S<2
E. 2<S<5/2
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Source: — Problem Solving |

by Anurag@Gurome » Wed Jan 25, 2012 8:08 pm
sud21 wrote:1/2+1/4+1/8+...+1/512=?
Intuitive Approach:
As we are always adding some values to 1/2, the sum is definitely going to be greater than 1/2. Now what are we doing here is adding half of the previous fraction to the sum. Hence, the sum is never going to reach 1.

For example, let us calculate sum of first few terms.
Sum of first 2 terms = (1/2 + 1/4) = 3/4 < 1
Sum of first 3 terms = (1/2 + 1/4 + 1/8) = 7/8 < 1
Sum of first 2 terms = (1/2 + 1/4 + 1/8 + 1/16) = 15/16 < 1
.. and so on

Mathematical Approach:
If we observe carefully the sum is nothing but a sum of geometric progression with common ratio, r = 1/2. As the common ratio is less than 1, we can find the sum of the geometric progression upto infinite number of terms.

Sum of infinite terms = r/(1 - r) = (1/2)/(1 - 1/2) = 1

Now, sum of the given number of terms must be less than the sum of infinite number of terms. Hence, the given sum must be less than 1.

The correct answer is B.
Anurag Mairal, Ph.D., MBA
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