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Assumed that i asked a question like 30 min ago thinking i got the hang of this, seems not.

So $$1^2+4^2+7^2+\dots+(3n-2)^2=\frac12n(6n^2-3n-1) \text{ for all } n\in\mathbb N$$ This time it seems way harder with the squares. so i did the steps and got stuck on the 3rd step(Again).

Step 1: prove LHS = RHS which it does for n=1

Step 2: Assume $n=k$ is true $$1^2+4^2+7^2+\dots+(3k-2)^2=\frac12k(6k^2-3k-1)$$

Step 3: would $n = k+1$? And would $n = k+1$ work for all equations?could someone help me with the last step, would be appreciated thanks

EDIT: Cheers for the help, i know where i went wrong!

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  • $\begingroup$ What have you tried so far for step3? Also just so you know, you can get nicely formatted fractions using \frac{}{}. $\endgroup$ – shuttle87 Dec 9 '13 at 23:37
  • $\begingroup$ The way you phrased step 3 makes me think your question might be a bit deeper than this. (For example, I'm not sure what "$n = k + 1$ work for all equations" means) You want to prove $P(n)$ is true for all $n$. In steps $2$ and $3$, what you want to do is show that, if $P(k)$ is true for some $k$, then $P(k+1)$ is also true. So, the proof part goes like "assume that $P(k)$ is true. do some algebra to it that shows that $P(k+1)$ is true". $\endgroup$ – Henry Swanson Dec 9 '13 at 23:47
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Hypothesis: $$ 1^2+4^2+7^2+\dots+(3n-2)^2=\frac{1}{2}n(6n^2-3n-1) $$

Thesis: $$ 1^2+4^2+7^2+\dots+(3n-2)^2+(3(n+1)-2)^2=\frac{1}{2}(n+1)(6(n+1)^2-3(n+1)-1) $$

By the induction hypothesis $$ 1^2+4^2+7^2+\dots+(3n-2)^2+(3(n+1)-2)^2= \frac{1}{2}n(6n^2-3n-1)+(3(n+1)-2)^2 $$ Write this as a polynomial in $n$; write $$ \frac{1}{2}(n+1)(6(n+1)^2-3(n+1)-1) $$ as a polynomial in $n$.

Verify that the expressions are the same. End.

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  • $\begingroup$ What happens to the (3(n+1)−2)^2 cause i got the exact same answer but confused on that appearing $\endgroup$ – Unknownstarz Dec 9 '13 at 23:45
  • $\begingroup$ @Unknownstarz Are you able to write $(3(n+1)-2)^2=(3n+1)^2=9n^2+6n+1$? I believe so. $\endgroup$ – egreg Dec 9 '13 at 23:48
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Well... $(3 - 2)^2 = 1 = \frac{6-3-1}{2}$ so it is true for $n=1$. Now, suppose it's true for $n=k$. Then, $1^2+4^2+\dots+(3k-2)^2+(3k+1)^2 = \frac{k\cdot(6k^2-3k-1)}{2} + (3k+1)^2 =\cdots$

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$$1^2+4^2+7^2+\dots+(3n-2)^2=\frac{1}{2}n(6n^2-3n-1), \forall n\in N$$

$$1^2+4^2+7^2+\dots+(3n-2)^2+(3n+1)^2=\frac{1}{2}n(6n^2-3n-1))+(3n+1)^2 $$ $$=\frac{1}{2}(6n^3-3n^2-n+2(3n+1)^2)=\frac{1}{2}(6n^3+15n^2+11n+2))= $$ $$\frac{1}{2}(6n^3+12n^2+6n+3n^2+5n+2)=$$ $$\frac{1}{2}(6n(n^2+2n+1)+3n^2+3n+2n+2)=$$ $$\frac{1}{2}(6n(n+1)^2+3n(n+1)+2(n+1))=$$ $$\frac{1}{2}(n+1)(6n(n+1)+3n+2)=\frac{1}{2}(n+1)(6n^2+9n+2)=$$ $$=\frac{1}{2}(n+1)(6n^2+12n+6-3n-3-1)=$$ $$=\frac{1}{2}(n+1)(6(n+1)^2-3(n+1)-1)$$

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