Prove that 1 + 4 + 7 … (3n-2) = ½(n)(3n-1) for all values of n e N.

This statement can also be written:

1) $\sum\limits_{n = 1}^k {(3n - 2)} = \frac{n}{2}(3n - 1)$

The formulaic response to proving the case for N is to write:

2) For n = k = 1, establish that both sides of 1) are equal.

a) 3k-2 = 3(1) – 2 = 1

b) (k/2)(3k-1) = ½(3(1)-1 = 1

c) Case proved for n = 1.

3) For n = k, assume that $\sum\limits_{n = 1}^k {(3k - 2)} = \frac{k}{2}(3k - 1)$ holds.

4) For k + 1, show that $\sum\limits_{n = 1}^{k + 1} {(3k - 2)} = \frac{{(k + 1)}}{2}(3(k + 1) - 1)$ holds.

5) $\sum\limits_{n = 1}^{k + 1} {(3k - 2)} = \left( {\sum\limits_{n = 1}^k {(3k - 2)} } \right) + (3k - 2)$ , substitution from 3) yields:

6) $\sum\limits_{n = 1}^{k + 1} {(3k - 2)} = \left( {\frac{k}{2}(3k - 1)} \right) + (3k - 2)$ , several steps of basic algebraic manipulation yield:

7) $\sum\limits_{n = 1}^{k + 1} {(3k - 2)} = \frac{{(k + 1)}}{2}(3(k + 1) - 1)$ thus it has been shown that by assuming statement 3) one can derive statement 4) thus demonstrating that equation 1) holds for k+1

8) Since 1) holds for n = k = 1 and holds for k+ 1 whenever it holds for k, then it holds for all n of N by P.M.I.

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Now I would like to take a different approach.

Recursion Theorem: Given a set X, a e X, and a function F:X-->X , there exists a unique function G:N-->X such that G(1) = a, and G(s(n)) = F(G(n)) for all n e N.

Prove the following statement is true for all N.

1) $\sum\limits_{n = 1}^k {(3n - 2)} = \frac{n}{2}(3n - 1)$

2) Let $L(k) = \sum\limits_{n = 1}^k {(3n - 2)} $

3) Let $R(k) = \frac{k}{2}(3k - 1)$

4) Showing that L(k) is well defined:

a) Let k = 1, then L(1) = (3(1)-2) = 1, L(k) is defined for k = 1.

b) Let F:N-->N, F(x) = x + (3n-2)

c) Then, since $L(k + 1) = F\left( {\sum\limits_{n = 1}^k {(3n - 2)} } \right) = \sum\limits_{n = 1}^{k + 1} {(3n - 2)} $ it is established (by the Recursion theorem) that L(k) is well defined over N_{1. }

5) Showing that R(k) is well defined:

a) Let k = 1, then (1/2) (3(1)-1) = 1, R(k) is defined for k = 1

b) Let F:N-->N, F(x) = x + (3n-2)

c) Then since $R(k + 1) = F\left( {\frac{k}{2}(3k - 1)} \right) = \left( {\frac{k}{2}(3k - 1)} \right) + (3k - 2) = \left( {\frac{{(k + 1)}}{2}(3(k + 1) - 1)} \right)$

it is established (by the Recursion theorem)that L(k) is well defined over N_{1 }

6) Finally, since the same function F was used to produce the “successoring” functions of L(k) and R(k) they must both (by the Recursion theorem) be generating the same set of images over N and hence each is not only well defined but equal to one another, therefore proposition 1) is proved to hold for all N_{1}.

My brain is still ping ponging over whether this is right. Certainly any g can be composed within a specific F but only the right g will allow F to identically reproduce it’s input sequence of images from g albeit offset by one unit increment.

So, the fact that, independently, F produces the unique successoring function for L AND R, and although both L and R appear very different from one another as do their respective successoring functions, both L and R produce the same set of pairs over N and are hence equal to one another, i.e identities of each other …. I think … my brain is still balking over this unlikely route of testing identities.

Also, I am not sure that I have all of the notation right.

Comments, corrections, clarifications most welcome. Thank you.