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Question 1203549: What is the largest number of consecutive odd positive integers that can be added together before the sum exceeds $40$?
Answer by math_tutor2020(3817)   (Show Source):
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One way to do it:
1+3 = 4
1+3+5 = 4+5 = 9
1+3+5+7 = 9+7 = 16
1+3+5+7+9 = 16+9 = 25
1+3+5+7+9+11 = 25+11 = 36
1+3+5+7+9+11+13 = 36+13 = 49
Adding the first 6 positive odd integers is the most we can do before exceeding 40.
Another approach
The sequence of positive odd integers {1,3,5,7,...} is arithmetic
a1 = 1 = first term of arithmetic sequence
d = 2 = common difference
Sn = sum of first n terms of arithmetic sequence
Sn = (n/2)*(2*a1+d*(n-1))
Sn = (n/2)*(2*1+2*(n-1))
Sn = (n/2)*(2n)
Sn = n^2
The sum of the first n positive odd integers is n^2
A few examples of this are shown in the previous section above.
We want to know when n^2 > 40
Apply the square root to both sides to get roughly n > 6.32
Summing the first 6 positive odd integers gets us 6^2 = 36
Summing the first 7 positive odd integers gets us 7^2 = 49
Answer: 6
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