MHB Factorization of an expression

[anemone]

Factorize the expression $(1+a+\cdots+a^n)^2-a^n$.

 

[Fermat1]

Factorize the expression $(1+a+\cdots+a^n)^2-a^n$.

Spoiler

$(1+..+a^n)^2=1+2a+..+(n+1)a^n+na^{n+1}+..+a^{2n}$.

So $(1+a+\cdots+a^n)^2-a^n=1+..+na^n+na^{n+1}+..+a^{2n}=(1+a+..+a^{n+1})(a^{n-1}+..+1)$

 

[anemone]

Spoiler

$(1+..+a^n)^2=1+2a+..+(n+1)a^n+na^{n+1}+..+a^{2n}$.

So $(1+a+\cdots+a^n)^2-a^n=1+..+na^n+na^{n+1}+..+a^{2n}=(1+a+..+a^{n+1})(a^{n-1}+..+1)$

Thanks for participating, Fermat. But I don't think it's straightforward to conclude that $1+..+na^n+na^{n+1}+..+a^{2n}$ is actually the product of $(1+a+..+a^{n+1})$ and $(a^{n-1}+..+1)$.

 

[kaliprasad]

let f(a) =$ (1+ a+ \cdots + a^n)^2 - a^n$

Spoiler

so f(a)(a-1)^2 = $(a^{n+1} -1)^2 - a^n(a-1)^2$
= $(a^{n+1}-1)^2 - a^n(a-1)^2$
= $a^{2n+2}- 2 a^{n+1} + 1 - a^n(a^2 - 2a + 1)$
= $a^{2n+2} - a^{n+2} - a^n + 1$
= $(a^{n+2} - 1)(a^n-1)$

so f(a)=$((a^{n+2} -1) / (a-1) * (a^n - 1)/(a-1)) $
= $ ( 1 + a + \cdots + a^{n+1}) ( 1 + a + \cdots + a^ {n-1})$

 

[Fermat1]

Thanks for participating, Fermat. But I don't think it's straightforward to conclude that $1+..+na^n+na^{n+1}+..+a^{2n}$ is actually the product of $(1+a+..+a^{n+1})$ and $(a^{n-1}+..+1)$.

No it's not straightforward, but it is suggested by the fact that the coefficients decrease after $a^{n+1}$.

 

[anemone]

No it's not straightforward, but it is suggested by the fact that the coefficients decrease after $a^{n+1}$.

Oh okay.