How to resolve it? How to find that sum?
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7Ask Gauss the great! :-) – Spock Jan 31 '14 at 18:00
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1Have you tried anything? Can we see your progress? – quapka Jan 31 '14 at 18:00
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This is a sum of arithmetic progression. – JohnK Jan 31 '14 at 18:01
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1+789999, 2+789998, 3+789997 etc. Good luck. – ir7 Jan 31 '14 at 18:05
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See Faulhaber's formulas. – Lucian Jan 31 '14 at 18:20
3 Answers
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If you can find the average of those numbers, then multiply that by $789999$, then you've got it.
The average of $1$ and $789999$ is $\dfrac{1+789999}{2}=395000$.
The average of $2$ and $789998$ is $\dfrac{2+789998}{2}=395000$.
The average of $3$ and $789997$ is $\dfrac{3+789997}{2}=395000$.
If you can figure out why the pattern persists, and write an explanation of it, then you should be able to see what the average of all the numbers from $1$ through $789999$ is.
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Hint: $$\sum_{k=0}^{10} k = 0 + 1 + 2 + 3 + 4 + + 6 + 7 + 8 + 9 + 10$$ $$= 0 + 10 + 1+9 + 2+ 8+ 3 + 7+4+6+ 5 $$ $$ = 5 \cdot 10 + 5 = 55= \frac{10(10+1)}{2}$$ $$\sum_{k=0}^{11} k = 0 + 1 + 2 + 3 + 4 + + 6 + 7 + 8 + 9 + 10 + 11 $$ $$= 11 +0 + 10 + 1+9 + 2+ 8+ 3 + 7+4+6+ 5 $$ $$ = 6 \cdot 11 = 66 = \frac{11(11+1)}{2}$$ etc
Can you see a pattern?
Nigel Overmars
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$$1+2+3+\cdots+n=\sum_{i=1}^ni=\dfrac {n(n+1)}2 \\ \ \\ \text{in your case: }1+2+3+\cdots+789999=\dfrac{789999\cdot790000}{2}=312049605000$$