Question

Does this seem legit?

Answer 1

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, \(1^2+2^2+...+n^2=\frac{1}{6}n(n+1)(2n+1)\). \(\blacksquare\)\[\]

Answer 2

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 \( n \geq 1 \).

Answer 3

will do, thank you! :)