How to reason inside expressions?

Question

In an expression like $$\sum_{k=1}^n (k^2-1)$$ I know (somehow, but how exactly is unclear) that $k$ is a real number, thus I can use all the applicable identities to rewrite it as $$\sum_{k=1}^n (k+1)(k-1)$$ But the reasoning should happen inside the expression, as $k$ is bound to the $\Sigma$, yet what I just did was outside of it. Casting this problem aside, I am not sure how I can use $k\in\mathbb R$. What indicates that it is a useable statement?

Question:

How to prove that

$$\sum_{k=1}^n (k^2-1)= \sum_{k=1}^n (k+1)(k-1)$$

Answer 1

If you set out carefully the definition of what $S_n=\sum_{k=1}^{n} a_k$ means it may help.

I think it is defined recursively: $$ S_1:=a_1\ \text{and then}\ \ S_{n+1}:=S_n + a_{n+1}. $$

You can now prove that whenever $a_k=b_k$ for all $k$ then $\sum_{k=1}^{n} a_k=\sum_{k=1}^{n} b_k$. (Yours is the special case $a_k=(k^2-1)$, $b_k=(k-1)(k+1)$.)

Base case: $$\sum_{k=1}^{1} a_k= a_1 =b_1 =\sum_{k=1}^{1} b_k,$$ the equalities justified by definition of $\sum_{1}^{1}$, hypothesis, definition of $\sum_{1}^{1}$.

Inductive step: $$\sum_{k=1}^{n+1} a_k= \sum_{k=1}^{n} a_k +a_{n+1}= \sum_{k=1}^{n} b_k +a_{n+1}= \sum_{k=1}^{n} b_k +b_{n+1}= \sum_{k=1}^{n+1} b_k, $$ where the equalities are justified by definition of $\sum_{1}^{n+1}$, inductive hypothesis, hypothesis, definition of $\sum_{1}^{n+1}$.

In terms of the way you explain your difficulty, note that we have used the definition of the $\Sigma$ symbol to "unbind" the top term, and then operate on it as we please, before using the definition of $\Sigma$ to "rebind" it.

Answer 2

The thing that will help you is one of these properties of equality:
1)If $$x=y$$ then for any function $f$ $$f(x)=f(y)$$
2)If $$x=y$$ then for any prediacte $\phi$, $\phi(x)<=>\phi(y)$
With either of this you can prove what you want to.
With (1) you can take function $f$ such that for any $x$ $$f(x)=\sum_{k=1}^n x $$ So $$k^2-1=(k-1)(k+1)$$ Applying first property:$$\sum_{k=1}^n k^2-1 = \sum_{k=1}^n (k-1)(k+1)$$
With (2) you can take predicate $\phi$ such that $\phi(x)$ means $$\sum_{k=1}^n x = \sum_{k=1}^n (k-1)(k+1)$$ So $$(k-1)(k+1)=k^2-1$$ So if $\phi((k-1)(k+1))$ is true then $\phi(k^2-1)$ is true. But $\phi((k-1)(k+1))$ is true by reflexive property of equality because $\phi((k-1)(k+1))$ is $$ \sum_{k=1}^n (k-1)(k+1) = \sum_{k=1}^n (k-1)(k+1)$$ So we must have $\phi(k^2-1)$ also be true, so $$\sum_{k=1}^n k^2-1 = \sum_{k=1}^n (k-1)(k+1)$$ is true.