.
The given sum is the sum of the first n positive odd integer numbers from 1 to 2n-1
S = 1 + 3 + 5 + . . . + (2n-1).
It is well known fact that this sum is equal to .
So, the problem asks to find "n" such that
> 4000.
Notice that = 63.25 (approximately).
Therefore, the required value of n is 64.
ANSWER. The least number of "n" is 64.
Solved.
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For introductory lessons on arithmetic progressions see
- Arithmetic progressions
- The proofs of the formulas for arithmetic progressions
- Problems on arithmetic progressions
- Word problems on arithmetic progressions
in this site.
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- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".
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Free of charge online textbook in ALGEBRA-II
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