find formula to sum all odd integers, from 0 up to 99, that user inputs
- Thread starter lyra002
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Please reply with the full and exact text of the original exercise (which would not have included zero as an odd integer), the complete instructions, and a listing of all formulas (summation, etc) that you've been given. When you reply, please include a clear statement of your thoughts and efforts so far, as this will greatly help us understand what you're needing to do and how you're (probably) expected to do it.
Thank you!
pka
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Hint: The sum you want is \(\displaystyle \sum\limits_{k = 1}^{50} {\left( {2k - 1} \right)} \)
pka
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It is perfectly clear if one wants students to rely on basics not memory.
\(\displaystyle \sum\limits_{k = 1}^{50} {\left( {2k - 1} \right)}=2\dfrac{(50)(51)}{2}-50=(50)^2 \)
HallsofIvy
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If you are not sure how to handle a general problem, it is always a good idea to try simple versions. The sum of the first 5 odd number is 1+ 3+ 5+ 7+ 9= 25. That easier to see if we write
1+ 3+ 5+ 7+ 9 and then
9+ 7+ 5+ 3+ 1 each vertical pair sums to 10 and there are 5 such sums- the total is 5(10)= 50. Since we added twice, the sum is 25.
More generally, if we write
1+ 3+ 5+ ...+n-4+n-2+ n
n+ n-2+n-4+ ...+ 5+ 3+ 1 each vertical pair sums to n+ 1 and there are (n+ 1)/2 terms (remember that n is odd)- the total is \(\displaystyle (n+1)(n+1)/2= (n+1)^2/2\). Since we added twice, the original sum is \(\displaystyle (n+1)^2/4\) (note that, since n is odd, this is an integer).
1+ 3+ 5+ 7+ 9 and then
9+ 7+ 5+ 3+ 1 each vertical pair sums to 10 and there are 5 such sums- the total is 5(10)= 50. Since we added twice, the sum is 25.
More generally, if we write
1+ 3+ 5+ ...+n-4+n-2+ n
n+ n-2+n-4+ ...+ 5+ 3+ 1 each vertical pair sums to n+ 1 and there are (n+ 1)/2 terms (remember that n is odd)- the total is \(\displaystyle (n+1)(n+1)/2= (n+1)^2/2\). Since we added twice, the original sum is \(\displaystyle (n+1)^2/4\) (note that, since n is odd, this is an integer).
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