"For the given statement Pn, write the statements P1, Pk, and Pk+1." 2 + 4 + 6 + . . . + 2n = n(n+1) Could someone show me how to do this problem?

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"For the given statement Pn, write the statements P1, Pk, and Pk+1." 2 + 4 + 6 + . . . + 2n = n(n+1) Could someone show me how to do this problem?

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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@ganeshie8 will you be able to help me?
P1-->n=1 .2=1*2 Pk-->n=k 2+4+6+...+2k=k(k+1) pK+1--> n=k+1 2+4+6+8+...+2k+2(k+1)=(k+1)(k+2) then we should show that, if the proposition is true for an arbitrary number k, then it must be true for the next number:k+1 2+4+6+...+2k+(2k+2)=k^2+k+2k+2=k^2+3k+2=(k+1)(k+2) so we proved that if we add 2(k+1) to HYPOTHESIS we can prove that 2+4+6+8+...+2k+2(k+1)=(k+1)(k+2)

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