1. **Problem statement:** Calculate the sum \( S = C_n^1 + 2C_n^2 + 3C_n^3 + \cdots + nC_n^n \). 2. **Formula and rules:** Recall the binomial theorem: $$ (1 + x)^n = \sum_{k=0}^n C_n^k x^k $$ 3. Differentiate both sides with respect to \( x \): $$ n(1 + x)^{n-1} = \sum_{k=0}^n k C_n^k x^{k-1} $$ 4. Multiply both sides by \( x \): $$ n x (1 + x)^{n-1} = \sum_{k=0}^n k C_n^k x^k $$ 5. Substitute \( x = 1 \) to find the sum: $$ n \cdot 1 \cdot (1 + 1)^{n-1} = \sum_{k=0}^n k C_n^k 1^k = \sum_{k=0}^n k C_n^k $$ 6. Since \( C_n^0 = 1 \) but multiplied by 0 in the sum, the sum starts effectively from \( k=1 \), so: $$ S = \sum_{k=1}^n k C_n^k = n 2^{n-1} $$ **Final answer:** $$ \boxed{C_n^1 + 2C_n^2 + 3C_n^3 + \cdots + nC_n^n = n 2^{n-1}} $$