Posted by Brock on Sunday, August 5, 2012 at 6:52pm.
Use mathematical induction to prove that the statement holds for all positive integers. Also, can you label the basis, hypothesis, and induction step in each problem. Thanks
1. 2+4+6+...+2n=n^2+n
2. 8+10+12+...+(2n+6)=n^2+7n

AP Calc  Steve, Monday, August 6, 2012 at 4:32am
assume true for n=k. Then when n=k+1, we have
2+4+...+2k+(2k+2) = k^2 + k + 2k+2
= k^2 + 2k + 1 + k + 1
= (k+1)^2 + (k+1)
Since true for n=1, true for n=2,3,4...
Similarly,
8+10+...+(2k+6)+(2k+8) = k^2 + 7k + (2k+8)
= k^2 + 2k + 1 + 7k + 7
= (k+1)^2 + 7(k+1)
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