Chứng minh đẳng thức :
           $n4^{n-1}C^0_n - (n-1)4^{n-2}C^1_n+...+(-1)^{n-1}C^{n-1}_n =  C^1_n+2^2C^2_n+...+n2^{n-1}C^n_n$
Ta có :
     $ (2x-1)^n=C^0_n(2x)^n-C^1_n(2x)^{n-1}+...+(-1)^nC^n_n.$             $ (1)$
Lấy đạo hàm theo $x$ hai vế của $(1)$ ta được :
      $2n(2x-1)^{n-1}=2nC^0_n(2x)^{n-1}-2(n-1)C^1_n(2x)^{n-2}+$
                                                      $ +...+(-1)^{n-1}2C^{n-1}_n.$         $ (2)$
Thay $ x=2$ vào $(2)$ ta được :
      $n3^{n-1}=n4^{n-1}C^0_n-(n-1)4^{n-2}C^1_n+...+(-1)^{n-1}C^{n-1}_n$.     $(3)$
Mặt khác, ta có :
      $(1+x)^n=C^0_n+C^1_nx+C^2_nx^2+C^3_nx^3+...+C^n_nx^n$          $(4)$
Lấy đạo hàm theo $x$ hai vế của $(4)$ ta được :
      $n(1+x)^{n-1}=C^1_n+2C^2_nx+...+nC^n_nx^{n-1}.$                       $ (5)$
Thay $x=2$ vào $(5)$ ta được :
      $n3^{n-1}=C^1_n+4C^2_n+...+n2^{n-1}C^n_n$                                 (6)
Từ (3) và (6) ta suy ra điều phải chứng minh
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