Ta có: $(1+x)^n=C_n^0+C_n^1x+\ldots+C_n^nx^n$Tích phân 2 vế ta được:
$\int\limits_0^\frac{1}{2}(1+x)^n=\int\limits_0^\frac{1}{2}\left(C_n^0x^n+C_n^1x^{n-1}+\ldots+C_n^n\right)$
$\Leftrightarrow \dfrac{1}{n+1}(1+x)^{n+1}\left|\begin{array}{l}\dfrac{1}{2}\\0\end{array}\right.=\left(C_n^0x+\dfrac{1}{2}C_n^1x^2+\ldots+\dfrac{1}{n+1}C_n^nx^{n+1}\right)\left|\begin{array}{l}\dfrac{1}{2}\\0\end{array}\right.$
$\Rightarrow C_n^0.\dfrac{1}{2}+\dfrac{1}{2}C_n^1.\dfrac{1}{2^2}+\ldots+\dfrac{1}{n+1}C_n^n.\dfrac{1}{2^{n+1}}=\dfrac{\left(\dfrac{3}{2}\right)^{n+1}-1}{n+1}$
$\Rightarrow C_n^02^n+\dfrac{1}{2}C_n^12^{n-1}+\ldots+\dfrac{1}{n+1}C_n^n=\dfrac{3^{n+1}-2^{n+1}}{n+1}$