Define $f(x)=-\frac{2x}{4x+3}$ and $g(x)=\frac{x+2}{2x+1}$. Moreover, let $h^{n+1} (x)=g(f(h^n(x)))$, where $h^1(x)=g(f(x))$. If the value of $\sum_{k=1}^{100} (-1)^k\cdot h^{100}(k)$ can be written in the form $ab^c$, for some integers $a,b,c$ where $c$ is as maximal as possible and $b\ne 1$, find $a+b+c$.

[i]Proposed by [b]AOPS12142015[/b][/i]
1. **Define the functions and initial conditions:**
   \[
   f(x) = -\frac{2x}{4x+3}, \quad g(x) = \frac{x+2}{2x+1}, \quad h^1(x) = g(f(x))
   \]

2. **Compute \( h^1(x) \):**
   \[
   h^1(x) = g(f(x)) = g\left(-\frac{2x}{4x+3}\right)
   \]
   Substitute \( y = f(x) = -\frac{2x}{4x+3} \) into \( g(y) \):
   \[
   g(y) = \frac{y+2}{2y+1}
   \]
   Therefore,
   \[
   h^1(x) = g\left(-\frac{2x}{4x+3}\right) = \frac{-\frac{2x}{4x+3} + 2}{2\left(-\frac{2x}{4x+3}\right) + 1}
   \]
   Simplify the numerator:
   \[
   -\frac{2x}{4x+3} + 2 = \frac{-2x + 2(4x+3)}{4x+3} = \frac{-2x + 8x + 6}{4x+3} = \frac{6x + 6}{4x+3}
   \]
   Simplify the denominator:
   \[
   2\left(-\frac{2x}{4x+3}\right) + 1 = \frac{-4x}{4x+3} + 1 = \frac{-4x + 4x + 3}{4x+3} = \frac{3}{4x+3}
   \]
   Thus,
   \[
   h^1(x) = \frac{\frac{6x + 6}{4x+3}}{\frac{3}{4x+3}} = \frac{6x + 6}{3} = 2(x + 1)
   \]

3. **Induction hypothesis:**
   Assume \( h^n(x) = 2^n x + m \) for some constant \( m \).

4. **Inductive step:**
   \[
   h^{n+1}(x) = g(f(h^n(x))) = g(f(2^n x + m))
   \]
   Since \( f(x) \) and \( g(x) \) are linear transformations, the form \( 2^n x + m \) will be preserved, and the constant term will not affect the alternating sum.

5. **Sum calculation:**
   \[
   \sum_{k=1}^{100} (-1)^k \cdot h^{100}(k) = \sum_{k=1}^{100} (-1)^k \cdot (2^{100} k + m)
   \]
   The constant term \( m \) cancels out in the alternating sum:
   \[
   \sum_{k=1}^{100} (-1)^k \cdot 2^{100} k = 2^{100} \sum_{k=1}^{100} (-1)^k k
   \]
   The sum of the alternating series:
   \[
   \sum_{k=1}^{100} (-1)^k k = (1 - 2) + (3 - 4) + \cdots + (99 - 100) = -50
   \]
   Therefore,
   \[
   2^{100} \cdot (-50) = -50 \cdot 2^{100}
   \]

6. **Express in the form \( ab^c \):**
   \[
   -50 \cdot 2^{100} = -1 \cdot 5 \cdot 2 \cdot 2^{100} = -1 \cdot 5 \cdot 2^{101}
   \]
   Here, \( a = -1 \), \( b = 2 \), and \( c = 101 \).

7. **Sum \( a + b + c \):**
   \[
   a + b + c = -1 + 2 + 101 = 102
   \]

The final answer is \( \boxed{102} \)