How solve the problem $f(x+2)=f(x)+4x+4$ for any $x$

Question

Find $f(2012)$, when $f(2)=0$ and
$$f(x+2)=f(x)+4x+4$$ for any $x$

I tried to find $f(4), f(6), f(8),....$

\begin{align*} f(4)&=f(2)+4 \cdot 2 +4 \\ f(6)&=f(4)+4 \cdot 4 + 4 = f(2)+4 \cdot (2 + 4) + 2 \cdot 4 \\ f(8) &= f(2) + 4 \cdot (2 + 4 + 6) + 3 \cdot 4 \\ \vdots \\ f(2012) &= f(2) + 4 \cdot (2 + 4 + 6 + … + 2010) + 1005 \cdot 4 = 4 048 140 \end{align*}

And it is correct answer.

But my question is how they found $f(2012)$ as

$$f(2012)=2012^2 - 4 = 4048140 \, ?$$

How they found proper function $?$

Answer 1

Given, $f(x+2)=f(x)+4x+4$

$$1. \quad{f(x+2+2)=f(x+2)+4 \cdot (x+2) +4\\ 2. \quad f(x+2+4){=\color{green}{f(x+4)}+4 \cdot (x+4) +4\\=\color{green}{f(x+2)+4 \cdot (x+2)+4}+4 \cdot (x+4) +4\\=f(x+2)+4\cdot(2x+2+4)+2\cdot4}\\}$$ By observations, $${\therefore f(x+2+2n){=f(x+2)+4\cdot\left(nx+\sum_{i=1}^{n}2i\right)+n\cdot4\\ =f(x+2)+4\cdot\left(nx+2\cdot\frac{n(n+1)}{2}\right)+n\cdot4}}$$

Put, $x=0$,$$f(2+2n){ =f(2)+4\cdot n(n+1)+n\cdot4\\ =4\cdot n(n+1)+n\cdot4\quad (\because f(2)=0)\\=4n^2+8n}$$

Now, put $2+2n=m$, $$f(m){=4\cdot\left(\frac{m-2}{2}\right)^2+8\cdot\frac{m-2}{2}\\ =(m-2)^2+4m-8\\ =m^2-4}$$

Hence, $f(2012)=2012^2-4=4048140$

Answer 2

Define $g(2x) = f(x)$, we find the difference sequence of $g$:

$$ 0 \quad 12 \quad 32 \quad 60 \quad ... \\ 12 \quad 20 \quad 28 \quad ... \\ 8 \quad 8 \quad ... \\\ 0 \quad ... $$

Observe that the $4$th row contains all zero. We then know that the sequence is determined entirely by the left diagonal. In fact,

$$g(n) = 12 \binom{n-1}{1} + 8 \binom{n-1}{2} = 4n^2-4$$

Answer 3

Given that $f(x+2)-f(x)=4x+4$

here you can make a telescopic sum

$$\require{cancel}\cancel{f(x+2)}- f(x)=4x+4$$ $$\require{cancel}\cancel{f(x+4)}-\require{cancel}\cancel{f(x+2)}=4x+12$$ $$\cancel{f(x+6)}-\cancel{f(x+4)}=4x+20$$ $$\cancel{f(x+8)}-\cancel{f(x+6)}=4x+28$$ $$\cancel{f(x+10)}-\cancel{f(x+8)}=4x+36$$ $$\cancel{f(x+12)}-\cancel{f(x+10)}=4x+44$$ $$.$$$$.$$$$.$$$$.$$$$f(x+2(n))-\cancel {f(x+2(n-1)})=4x+8n-4$$ because Right hand side is an A.P

the sum becomes

$$f(x+2n)-f(x)=4nx+4n^2=4n(x+n)\tag1$$

put $x=2$ in equation $1$

$$f(2n+2)-0=4n(n+2)\tag2$$

replace $n$ to $n-1$ in equation $2$

$$f(2(n-1)+2)-0=4(n-1)((n-1))+2)$$

hence

$$f(2n)=4(n^2)-4=(2n)^2-4$$

replace $2n$ by $x$ again

$$f(x)=x^2-4$$

Which was your intended doubt to the solution

Answer 4

One way to solve is through the unilateral $\mathcal{Z}$-Transformation. The $\mathcal{Z}$-Transform is linear and $\mathcal{Z}_{x}\left[ f\left( x + n \right) \right]\left( s \right) = s^{n} \cdot \left( \hat{f}\left( s \right) - \sum\limits_{k = 0}^{n - 1}\left[ f\left( k \right) \cdot x^{-k} \right]\right)$ applies ($\mathcal{Z}_{x}\left[ f\left( x \right) \right]\left( s\right) =: \hat{f}\left( s \right) \wedge n \in \mathbb{N}$). So: $$ \begin{align*} f\left( x + 2 \right) &= f\left( x \right) + 4 \cdot x + 4\\ \mathcal{Z}_{x}\left[ f\left( x + 2 \right) \right]\left( s \right) &= \mathcal{Z}_{x}\left[ f\left( x \right) + 4 \cdot x + 4 \right]\left( s \right)\\ \mathcal{Z}_{x}\left[ f\left( x + 2 \right) \right]\left( s \right) &= \mathcal{Z}_{x}\left[ f\left( x \right) \right]\left( s \right) + \mathcal{Z}_{x}\left[ 4 \cdot x + 4 \right]\left( s \right)\\ s^{2} \cdot \mathcal{Z}_{x}\left[ f \left( x \right) \right]\left( s \right) - f(0) \cdot s^{2} - f(1) \cdot s &= \mathcal{Z}_{x}\left[ f\left( x \right) \right]\left( s \right) + \frac{4 \cdot s^{2}}{\left( s - 1 \right)^{2}}\\ \mathcal{Z}_{x}\left[ f \left( x \right) \right]\left( s \right) &= \frac{\frac{4 \cdot s^{2}}{\left( s - 1 \right)^{2}} + f(0) \cdot s^{2} + f(1) \cdot s}{s^{2} - 1}\\ \end{align*} $$ $$\fbox{$ \begin{align*} f \left( x \right) &= \mathcal{Z}_{s}^{-1}\left[\frac{\frac{4 \cdot s^{2}}{\left( s - 1 \right)^{2}} + f(0) \cdot s^{2} + f(1) \cdot s}{s^{2} - 1} \right]\left( x \right)\\ f \left( x \right) &= \frac{2 \cdot x^{2} + \left( -1 \right)^{x} - 1 + f\left( 0 \right) \cdot \left( \left( -1 \right)^{x} + 1 \right) + f\left( 1 \right) \cdot \left( \left( -1 \right)^{x + 1} + 1 \right)}{2}\\ \end{align*} $}$$ (You can see it here too.)

$f\left( 1 \right)$ and $f\left( 0 \right)$ are some arbitrary constants. We get $f\left( 2 \right) = f\left( 0 \right) + 4 \Longleftrightarrow 0 = f\left( 0 \right) + 4 \Longleftrightarrow f\left( 0 \right) = -4$. $$\fbox{$\fbox{$ \begin{align*} f\left( x \right) &= \frac{2 \cdot x^{2} + \left( -1 \right)^{x} - 1 - 4 \cdot \left( \left( -1 \right)^{x} + 1 \right) + f\left( 1 \right) \cdot \left( \left( -1 \right)^{x + 1} + 1 \right)}{2}\\ \end{align*} $}$}$$

If $x = 2012$ we would get: $$ \begin{align*} f\left( 2012 \right) &= \frac{2 \cdot 2012^{2} + \left( -1 \right)^{2012} - 1 - 4 \cdot \left( \left( -1 \right)^{2012} + 1 \right) + f\left( 1 \right) \cdot \left( \left( -1 \right)^{2012 + 1} + 1 \right)}{2}\\ f\left( 2012 \right) &= \frac{2 \cdot 2012^{2} + 1 - 1 - 4 \cdot \left( 1 + 1 \right) + f\left( 1 \right) \cdot \left( -1 + 1 \right)}{2}\\ f\left( 2012 \right) &= \frac{2 \cdot 2012^{2} - 4 \cdot 2}{2}\\ f\left( 2012 \right) &= 2012^{2} - 4 = 4048140\\ \end{align*} $$ (You can see this here too.)