For $n=1$ its easy to see that the area of triangle $C_{1}B_{1}C_{2}$ is a half of the area of $\triangle ABC$. Now, we use induction to conclude: Let for $n=k$ the shaded area is $\frac{1}{2}$. Now for $n=k+1$ the shaded area is $(S_{\triangle C_{1}B_{1}C_{2}}+\cdots+S_{\triangle C_{k}B_{k}C_{k+1}})+S_{\triangle C_{k+1}B_{k+1}C_{k+2}}=\frac{1}{2}S_{\triangle AB_{k}C_{k+1}}+S_{\triangle C_{k+1}B_{k+1}C_{k+2}}=\frac{1}{2}\frac{k}{k+1}\frac{k+1}{k+2}S_{\triangle ABC}+\frac{1}{k+2}S_{\triangle ABC}=\frac{k}{2k+4}S_{\triangle ABC}+\frac{1}{k+2}S_{\triangle ABC} = \frac{1}{2}S_{\triangle ABC}$ and so we have completed the induction step and the answer is that the shaded area has $\frac{1}{2}$ of the area of triangle $\triangle ABC$.

Straightforward, we have that $S_{\triangle C_{i}B_{i}C{i+1}}=(\frac{1}{i+1})(\frac{i}{n})(\frac{i+1}{n+1})S_{\triangle ABC}=\frac{i}{n(n+1)}S_{\triangle ABC}$ Now we use the formula: $1+2+\cdots+n=\frac{n(n+1)}{2}$ : $$\sum\limits_{i=1}^{n} S_{\triangle C_{i}B_{i}C_{i+1}} = \sum\limits_{i=1}^{n} \frac{i}{n(n+1)}S_{\triangle ABC} = \frac{1}{2} S_{\triangle ABC} $$That means that the shaded area is half of the area of $\triangle ABC$.

From the second solution we know that $S_{C_{i}B_{i}C_{i+1}}=\frac{i}{n(n+1)}S_{\triangle ABC}$. On the other hand, if we look at $S_{B_{i}C_{i+1}B_{i+1}}=(\frac{1}{i+1})(\frac{i+1}{n})(\frac{i+1}{n+1})S_{\triangle ABC}=\frac{i+1}{n(n+1)}S_{\triangle ABC}$. That means that $S_{C_{i+1}B_{i+1}C_{i+2}}=S_{B_{i}C_{i+1}B_{i+1}}$. Now we are almost done: $$\sum\limits_{i=0}^{n-1} S_{C_{i+1}B_{i+1}C_{i+2}} = \frac{1}{2}\sum\limits_{i=0}^{n-1} S_{C_{i+1}B_{i+1}C_{i+2}} + \frac{1}{2}\sum\limits_{i=0}^{n-1} S_{B_{i}C_{i+1}B_{i+1}}= \frac{1}{2}S_{\triangle ABC}$$The second equality is derived from the definition of the triangles, which cover up the entire $\triangle ABC$.