Given finite summation, find suitable function Which function(s) $f(x)$ fulfill the following condition:
$\sum_{i=0}^n f(i) = 1 + \frac{f(0)}{2}$
If an additional condition is necessary, then the function can be restricted to any decreasing monotonic function. 
(If this is an existing homework question to any course, I'd be glad to know which course it is.)
 A: This property is way too weak if you're only speaking about one specific $n$. Given any fixed $n$, we must have
$$\sum_{i=0}^nf(i)=1+\frac12f(0)$$
which rewrites to
$$\sum_{i=1}^nf(i)=1-\frac12f(0)$$
Take any function, and set for example 
$$f(0)=2-2\sum_{i=1}^nf(i)$$
to make sure the property holds.
Even if we take into account that $f$ is a decreasing monotonic function (may I add, decreasing functions are always monotonic), there's still way too many functions to say anything useful about this.

Edit:
if we know $f(x)=ax^2+bx+c$ for some $a,b,c$, then we know with Faulhaber's formula
that
$$\sum_{i=1}^nai^2+bi+c=a\frac{2n^3+3n^2+n}{6}+b\frac{n^2+n}{2}+cn$$
and since $1+\frac{1}{2}f(0)=1+\frac c2$, we must solve (for $a,b,c$)

$$a\frac{2n^3+3n^2+n}{6}+b\frac{n^2+n}{2}+cn=1-\frac c2$$
or
$$a\frac{2n^3+3n^2+n}{6}+b\frac{n^2+n}{2}+c(n+\frac 12)=1$$
and if $n$ is fixed, then we can simply choose two of $a,b,c$ and determine what the last one has to be - if however $a,b,c$ are fixed, then we can solve this equation (for it's only cubic in $n$) but, depending on $a,b,c$ of course, this most likely gives you non-natural $n$ and thus would not be possible.
