MHB Prove f(n) is a product of two consecutive positive integers for all n

[Albert1]

$f(n)=\underbrace{111--1}\underbrace{222--2}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n$
prove:$f(n)$ is a product of two consecutive positive integers for all $n\in N$

 

[kaliprasad]

$f(n)=\underbrace{111--1}\underbrace{222--2}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n$
prove:$f(n)$ is a product of two consecutive positive integers for all $n\in N$

Spoiler

as n consecutive 1's is $\frac{10^n-1}{9}$ using GP
we have $f(n) = 10^n(\frac{10^{n}-1}{9} + 2 * (\frac{10^{n}-1}{9})$
$= \frac{10^n(10^{n} - 1) + 2(10^n-1)}{9}$
$= \frac{10^{2n} + 10^n -2}{9}$
$=\frac{(10^n-1)(10^n+2)}{9}$
= $(\frac{10^n-1}{3})(\frac{10^n+2}{3})$
= $(\frac{10^n-1}{3})(\frac{10^n-1}{3}+1)$
now $10^n$ leaves a raminder 1 when divided by 3 so $10^n-1$ is divsible by 3 and hence noth the terms above are
integers and difference is one.