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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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Show that \(1^2+2^2+...+n^2=\frac{1}{6}n(n+1)(2n+1)\). Proof: For \(n=1\), \(1^2+2^2+...+n^2=\frac{1}{6}n(n+1)(2n+1)\implies 1^2=\frac{1}{6}(2)(3)\implies1=1\), which is true. Assume that \(1^2+2^2+...+k^2=\frac{1}{6}k(k+1)(2k+1)\) is true, and show that \(1^2+2^2+...+k^2+(k+1)^2=\frac{1}{6}k(k+1)(2k+1)+(k+1)^2\) is also true. We have \(\frac{1}{6}k(k+1)(2k+1)+(k+1)^2\), \(\frac{1}{6}(k+1)(k+2)(2k+3)\), \(\frac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)\), which is true. Therefore, \(1^2+2^2+...+n^2=\frac{1}{6}n(n+1)(2n+1)\). \(\blacksquare\)\[\]
Yes, with some small changes in language. Let P(n) be the statement that 1^2 + 2^2 + ... + n^2 = stuff. You've shown P(1) is true, check. Now you want to show for all k greater than or equal to 1, P(k) => P(k+1) It's not that you start with P(k) is true. It's you just want to prove the implication. Then at the end, the best way to write this is summarize the situation. "We have now shown that P(1) is true and that for all \( k \geq 1 \) \[ P(k) \implies P(k+1) \] Therefore by the Principle of Mathematical Induction, we can conclude that \[ P(n) \] is true for all \( n \geq 1 \).
will do, thank you! :)

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