Question

Chứng minh rằng: \(C^1_n+2C^2_n+3C^3_n+...+nC^n_n=n.2^{n-1}\)

Answer 1

Với k \(\in\)N* ; ta có : \(kC_n^k=k.\dfrac{n!}{\left(n-k\right)!k!}=\dfrac{n!}{\left(n-k\right)!\left(k-1\right)!}=\dfrac{n\left(n-1\right)!}{\left[n-1-\left(k-1\right)\right]!\left(k-1\right)!}=nC_{n-1}^{k-1}\)

Khi đó : \(C_n^1+2C_n^2+...+nC^n_n\)  = \(\Sigma^n_{k=1}nC^{k-1}_{n-1}\)  

= \(n\left(C_{n-1}^0+C_{n-1}^1+...+C_{n-1}^{n-1}\right)\)  \(=n.\left(1+1\right)^{n-1}=n.2^{n-1}\) ( đpcm )