Sum of Consecutive Functions

The above equation can be rearranged as $f(x+ 1) = x - f(x)$

Then, $f(2013) = 2012 - f(2012)$

Similarly, $f(2012) = 2011 - f(2011)$

Substituting this value of $f(2012)$ in above equation gives

$f(2013) = 1 + f(2011)$

Since $f(2011) = 1 + f(2009)$ ,

$f(2013) = 2 + f(2009)$

$f(2013) = 3 + f(2007)$

$f(2013) = 4 + f(2005)$

$f(2013) = 5 + f(2003)$

$f(2013) = 6 + f(2001)$

$\phantom{abc} \vdots$

$f(2013) = 1006 + f(1)$

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Substituting this value in the question expression

$\frac{f(0) + f(1) + 1006}{2}$

However, $f(0) + f(1) = 0$

So, the answer is $\frac{1006}{2}$

$=\boxed{503}$

The given equation is $f(x) + f(x + 1) = x$ .

• Put x = 0,

$f(0) + f(1) = 0$

$f(1) = 0 - f(0)$

• Put x = 1,

$f(1) + f(2) = 1$

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

$f(2) = 1 + f(0)$

• Put x = 2,

$f(2) + f(3) = 2$

$f(3) = 2 - f(2) = 2 - 1 - f(0) = 1 - f(0)$

$f(3) = 1 - f(0)$

• Put x = 3,

$f(3) + f(4) = 3$

$f(4) = 3 - f(3) = 3 - 1 + f(0) = 2 + f(0)$

$f(4) = 2 + f(0)$

• Put x = 4,

$f(4) + f(5) = 4$

$f(5) = 4 - f(4) = 4 - 2 - f(0) = 2 - f(0)$

$f(5) = 2 - f(0)$

• From this, we can generalize for any value of n,

Here it is:

When n is odd, $f(n) = \frac{(n - 1)}{2} - f(0)$

When n is even, $f(n) = \frac{(n)}{2} + f(0)$

So,

• $f(2013) = \frac{(2013 - 1)}{2} - f(0) = \frac{(2012)}{2} - f(0) = 1006 - f(0)$

Hence,

• $\frac{f(0) + f(2013)}{2} = \frac{f(0) + 1006 - f(0)}{2} = \frac{1006}{2} = 503$

That's the answer!

• $f(0)+f(1)+f(2)+f(3)+...+f(2012)+f(2013)=0+2+...+2012= 1006\times 1007$ (1)
• $f(1)+f(2)+f(3)+f(4)+...+f(2011)+f(2012)=1+3+....+2011= 1006^2$ (2) subtracting (1) with (2) :
• $f(0)+f(2012)= 1006$
• $\frac{f(0)+f(2013)}{2}=503$

Given $f(x)+f(x+1)=x$ , Then $f(x+1)=x-f(x)$ *use this formula *

$f(2013)$

$=f(2012+1)$

$=2012-f(2012)$

$=2012-2011+f(2011)$

$=1+f(2011)$ <-----

$=1+2010-f(2010)$

$=1+2010-2009+f(2009)$

$=1+1+f(2009)$ <----- if $f(n)$ minus 2 ,will plus 1.

Because $2013$ minus $1006 \times 2$ left $1$ .So that ,will plus $1006$ .

At last $f(2013)$ will become $1006+f(1)$

Then, $f(1)=0-f(0)$

Substitute to the question:

$\frac{f(0)+1006-f(0)}{2}$

$=\frac{1006}{2}$

$\boxed{503}$

f(2x+1)+f(0)=x -> f(2*1006 + 1) + f(0) = 1006 -> [f(2013)+f(0)]/2=1006/2=503

First you must observe that

$f(2013) = f(2012 + 1) = (2012 - 2011) + (2010 - 2009) + \cdots + (2 - 1) + (0 - f(0))$

Also observe that we have $2012/2$ subtractions in parenthesis until $(2 - 1)$ (inclusive). All those subtractions have value $1$ so that

$\frac{f(0) + f(2013)}{2} = \frac{f(0) + \frac{2012}{2} - f(0)}{2} = \frac{2012}{4} = 503$

putting x=0 f(0)+f(1)=0 putting x=1 and multiplying whole expression by -1 -f(1)-f(2)=-1 then x=2 f(2)+f(3)=2 then x=3 and multiplying whole expression by -1 -f(3)-f(4)=-3 similarly following this putting of x and multiplying whole expression by -1 when x is odd

in end we get f(1)+f(2013)=0-1+2-3+4.......-2011+2012 = 1006

so (f(1)+f(2013))/2 =503

Phew. This is going to be lengthy.

When the question mentioned "sum of consecutive functions", I immediately started to figure out what kind of pattern does it have. I started out with a few questions:

1. $f(x) + f(x+1)=x$ is true if both the number are consecutive to each other. For example, $f(50) + f(51)=50$

2. However, if we try to find $f(x) + f(n)$ , it would be impossible with reasoning itself (from the best of my capability)

3. Hence, I begin looking for a pattern that allows the operation $f(x) + f(n)$ to be able to find its answer easily.

We want to find the general formula for $f(x) + f(n)$ that satisfies all the values of $x$ and $n$ , where $x$ and $n$ are not consecutive/adjacent to each other.

Let us start with $x=0$ and $n=1$

$f(0) + f(1) = 0$ -- (1)

$f(1) + f(2) = 1$ -- (2)

Hence, we want (1) to have the function f(2) in it.

From (2),

$f(1) = 1-f(2)$

So, $f(0) + (1-f(2))=0$

$f(0) - f(2) = -1$

Weird enough, -1 is the difference of f(0) and f(2). That is not what we're hoping for. But,

$f(2) + f(3) = 2$

$f(2) = 2-f(3)$

Hence,

$f(0) - (2-f(3)) = -1$

$f(0) + f(3) = 1$

We see that it becomes a sum when the next number comes in. Using the same method to insert the next function,

$f(0) - f(4) = -2$

$f(0) + f(5) = 2$

$f(0) - f(6) = -3$

$f(0) + f(7) = 3$

We can see a pattern where the result increases by one for every consecutive odd integer. This applies to even integers as well, except that they are negative values instead.

Hence,

$f(x) + f(n) = \frac{n-1}{2}$ for n = odd integer when x = 0

Of course, this equation is only true for x=0.

Hence, taking n = 2013,

$f(x) + f(2013) = \frac{2013-1}{2} = 1006$

Giving us 503 when halved again.

$It's\quad easy\quad to\quad show\quad that\quad the\quad function\quad is\quad :\\ \quad f(x)\quad =\quad \frac { 1 }{ 2 } x\quad -\quad \frac { 1 }{ 4 } \\$

assume x=0, f(0)+f(1)=0 ------ (1) x=1, f(1)+f(2)=1 ------(2) .
. . . x=2012, f(2012)+f(2013)=2012 ------(2013) the value could be obtained by solving as follows.. (1)-(2)+ ..........+(2013) f(0)+f(2013)=0-1+2-3+4-5+..........+2012 =(2+4+6+8+.......+2012)-(1+3+5+7+9+.....2011) =[(1006)(1006)]+1006-[(1006)(1006)] =1006 therefore, the value of [f(0)+f(2013)]/2 is 503 thank you

f(2013)= 2012-f(2012)= 2012-{2011-f(2011)}=1+f(2011)=.......=2+f(2009)=........=3+f(2007)=....=1006+f(1)=1006+0-f(0) =>{f(o)+f(2013)}/2=503

f(0)+f(1)=1. f(0)=-f(1). f(1)+f(2)=1. f(1)=1-f(2)................................ Then we get f(0)=1006-f(2013). Then( f(0)+f(2013))/2=1006/2=503

We can solve this problem with telescoping pattern, we have f(0) + f(1) = 0 ; f(1) + f(2) = 1; f(2) + f(3) = 2 ; ... ; f(2012) + f(2013) = 2012. Next, we sum(+/- , "telescoping pattern" ) all of statements above ---> f(0) + f(1) f(1) + f(2) f(2) + f(3) + ... - f(2012) + f(2012) + f(2013) = (0 - 1 + 2 - 3 + ... - 2011 + 2012)/2 = 1*1006/2 = 503. Answer : 503

$f(0)$ + $f(1)$ = 0

$f(1)$ + $f(2)$ = 1 $=>$ $f(1)$ = 1- $f(2)$

$f(2)$ = 2- $f(3)$

$f(3)$ = 3- $f(4)$

...

$f(0)$ + $f(1)$ =0 $<=>$ $f(0)$ +1-2+3-4+5-...+2011-2012+ $f(2013)$ =0

$f(0)$ + $f(2013)$ =2012-2011+2010-2009+...+2-1 = $\frac{2012}{2}$ = 1006

$\frac{f(0)+f(2013)}{2}=503$