For the given statement Pn, write the statements P1, Pk, and Pk+1. (2 points) 2 + 4 + 6 + . . . + 2n = n(n+1) Would the answer just be: (k+1)(k+2) ?

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For the given statement Pn, write the statements P1, Pk, and Pk+1. (2 points) 2 + 4 + 6 + . . . + 2n = n(n+1) Would the answer just be: (k+1)(k+2) ?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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well that is the sum of k + 1 terms....
the left side contains the terms including the general term 2n the right side contains the sum of the terms n(n + 1) so n = 1 term 1 = 2 the sum of 1 term 1(1 + 1) 2 the kth term 2k the sum of k terms is k(k + 1) the k + 1 term 2(k + 1) the sum is (k +1)(k + 2) this seems a lot like mathematical induction.
Yes, it is. But what should the final answer even look like? I'm confused by the problem.

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ok... so its true for n = 1 assume that for n = k the sum is k(k +1) now for the k + 1 term term k+ 1 = 2(k + 1) or 2k+ 2 if you add this to the sum of k terms you should get (k + 1)(k + 2) so sum of term k + 1 and the sum of k terms 2(k + 1) + k(k + 1) both terms have a common factor of (k + 1) so it can be written as (k + 1)(2 + k) or (k + 1)(k + 2) (1) now using the sum n(n + 1) for k + 1 terms it becomes (k + 1)(k + 1+1) or (k + 1)(k+2) so you have shown that the sum of k terms and the k + 1 term is equal to the sum of k +1 terms.. hope that makes sense
So what I wrote is correct then?
yes...
Ok thanks so much!

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