The value of ${}^{100}C_0+{}^{100}C_3+{}^{100}C_6+{}^{100}C_9+\mathrm{.....}{}^{100}C_{99}$ is $\frac{2^{k_1}-1}{k_2}$, then the sum of digits of   $(k_1+k_2)$ is equal to

$$\begin{gathered} {(1+x)}^{100}={}^{100}C_0+{}^{100}C_1{x}^1+{}^{100}C_2{x}^2\mathrm{......}{}^{100}C_{100}{x}^{100} \\ {(1+1)}^{100}={}^{100}C_0+{}^{100}C_1+{}^{100}C_2+{}^{100}C_3\mathrm{.......}+{}^{100}C_{100} \\ {(1+\omega )}^{100}={}^{100}C_0+{}^{100}C_1\omega +{}^{100}C_2{\omega }^2\mathrm{.......}+{}^{100}C_{100}\omega \\ {(1+{\omega }^2)}^{100}={}^{100}C_0+{}^{100}C_1{\omega }^2+{}^{100}C_2\omega \mathrm{.......}{}^{100}C_{100}{\omega }^2 \\ {2}^{100}+{\omega }^{200}+{\omega }^{100}=3\{{}^{100}C_0+{}^{100}C_3+{}^{100}C_6+{}^{100}C_9\mathrm{......}{}^{100}C_{99}\} \\ \frac{2^{100}-1}{3}={}^{100}C_0+{}^{100}C_3+{}^{100}C_6+{}^{100}C_9\mathrm{.......}{}^{100}C_{99} \end{gathered}$$