How do you find the sum of the first 1000 positive even integers?

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Answer 1 · 2016-10-16T07:37:16

An even integer number #n# can be written as #n=2* k#, with #k# an integer

The sum you are looking for is:

#2+4+6+8+...+2000#, but taking 2 as a common factor this is the same as:

#2(1+2+3+4+...+1000)#

Now we can use the formula for the sum:

#sum_(n=1)^1000 n=(1000*1001)/2=500*1001=500500#

Using this now we have:

#2(1+2+3+4+...+1000)=2*sum_(n=1)^1000 n=(1000*1001)/2=#

#=2*500500=1001000#