Asked by Brock
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
1. 2+4+6+...+2n=n^2+n
2. 8+10+12+...+(2n+6)=n^2+7n
Answers
Answered by
Steve
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)
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)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.