Question 650517: The sum of eight consecutive even integers; n = first integer of the eight
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! Let n be the first; (n+2) be the second...
Notice that for each kth even number, we can write it as (n+2(k-1)). This is unnecessary to solve the problem, but it may help you see where I come up with this next step:
So continuing on, the 8th must be (n+14)
So we have n+(n+2)+(n+4)+(n+6)+(n+8)+(n+10)+(n+12)+(n+14)
=8n + 2*(sum of 1 to 7)
=8n + 2*(28)
=8n+56
=8(n+7)
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Aside:
Curiously, what if it was the first 9 even numbers?
Then we'd have an added on (n+16).
This would be 9n + 2*(sum of 1 to 8)
= 9n + 2*(36)
= 9n + 72
= 9(n+8)
So, in general, the sum of k consecutive integers
= k(n+(k-1))
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