Pls explain the calculation

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by Stuart@KaplanGMAT » Fri Dec 20, 2013 12:51 am
Hi!

This question is testing your ability to find limits and estimate.

We're told that K is the sum of the reciprocals of the integers 43 through 48, inclusive. In other words:

K = 1/43 + 1/44 + 1/45 + 1/46 + 1/47 + 1/48

well, that's a whole mess of ugly math, so there's got to be a way to answer the question without actually calculating.

The quickest way to do so is to think of 2 extreme ranges and then know the answer has to be in the middle.

If we were summing 1/48 + 1/48 + 1/48 + 1/48 + 1/48 + 1/48, then the answer would be 6/48 = 1/8.

So, we know the answer has to be more than 1/8, since 5 of our numbers are bigger than the ones used in the estimate.

If we were summing 1/42 + 1/42 + 1/42 + 1/42 + 1/42 + 1/42, then the answer would be 6/42 = 1/7.

So, we know the answer has to be less than 1/7, since all 6 of our numbers are smaller than the ones used in the estimate.

Therefore, 1/8 < K < 1/7

Since 1/7 isn't there (and it would never be in the answers for this question, since you'd have to do the math to be 100% sure whether K were closer to 1/7 or 1/8), of the choices provided K is definitely closest to 1/8... choose (C)!
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by Mathsbuddy » Fri Dec 20, 2013 6:54 am
K = 1/43 + 1/44 + 1/45 + 1/46 + 1/47 + 1/48
Let N = 43
K = 1/N + 1/(N+1) + 1/(N+2) + 1/(N+3) + 1/(N+4) + 1/(N+5)
Using Notation 012345 = N(N+1)(N+2)(N+3)(N+4)(N+5)
then:

K = (12345 + 02345 + 01345 + 01245 + 01235 + 01234)/012345

The highest order of N in the solution will be:
5N^4/N^5 = 5/N = 5/43 ≈ 1/8
Choose C