evalute 1 +2-3+4+5-6.......+2005+2005+2006-2007+2008+2009
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It is the sum of all the first 2009 natural numbers - twice of all the multiples of 3 till 2007.
We know that sum of all natural numbers = n(n+1)/2
Therefore, we have,
2009*(2010)/2 - 2(3(669*670)/2) Because 3*669=2007.
We get, 2019045-1344690=674355.
We know that sum of all natural numbers = n(n+1)/2
Therefore, we have,
2009*(2010)/2 - 2(3(669*670)/2) Because 3*669=2007.
We get, 2019045-1344690=674355.
Answered by
1
Answer:1346700
Step-by-step explanation:
This is the correct answer
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