If S_{1}=\sum_{r=0}^{n}{ }^{n} C_{r}(-\frac{1}{2})^{r} ; S_{2}=\sum_{r=0}^{n}{ }^{n} C_{r}(-\frac{3}{4})^{r} . . . . . . S_{m}=\sum_{r=0}^{n}{ }^{n} C_{r}(-( {2^{m}-1}/{2^{m}}))^{r} then:

\sum_{k=1}^{4} S_{k}= {2^{4 n}-1}/{2^{4 n}(2^{n}-1)}

\sum_{k=1}^{16}(S_{k})^{\frac{1}{n}}=1-\frac{1}{2^{16}}

\sum_{k=1}^{4} S_{k}=1-\frac{1}{2^{16}}