ĐK của pt là \(n\ge2\)

\(\left(1+x\right)^n=C_n^0+x.C_n^1+x^2.C_n^2+x^3.C^3_n+x^4.C_n^4+...+x^n.C_n^n\)

\(\Rightarrow n\left(1+x\right)^{n-1}=C_n^1+2x.C_n^2+3x^2.C^3_n+4x^3.C_n^4...+n.x^{n-1}.C^n_n\) ( đạo hàm hai vế )

\(\Rightarrow n\left(n-1\right)\left(x+1\right)^{n-2}=2.C_n^2+2.3.x.C_n^3+3.4.x^2.C_n^4+...+\left(n-1\right)n.x^{n-2}.C_n^n\) ( đạo hàm hai vế )

Thay x=1 ta được: \(n\left(n-1\right).2^{n-2}=2.C_n^2+2.3.C^3_n+3.4.C_n^4+...+\left(n-1\right).n.C^n_n\)

\(\Leftrightarrow n\left(n-1\right).2^{n-2}=64n.\left(n-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}n\left(n-1\right)=0\\2^{n-2}=64\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=0;n=1\left(ktm\right)\\n=8\left(tm\right)\end{matrix}\right.\)

Vậy \(n=8\)

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 )

Ta có:

\(k.C_n^k=k.\dfrac{n!}{\left(n-k\right)!.k!}=n.\dfrac{\left(n-1\right)!}{\left(n-1-\left(k-1\right)\right)!\left(k-1\right)!}=n.C_{n-1}^{k-1}\)

Do đó:

\(1C_n^1+2C_n^2+...+nC_n^n\)

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

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

\(=n.2^{n-1}\)

\(\Leftrightarrow\dfrac{n!}{\left(n-3\right)!\cdot3!}+2n=\dfrac{n!}{\left(n-2\right)!}+1\)

\(\Leftrightarrow\dfrac{n\left(n-1\right)\left(n-2\right)}{6}+2n=\dfrac{\left(n-1\right)\cdot n}{1}+1\)

\(\Leftrightarrow n\left(n-1\right)\left(n-2\right)+12n=6n\left(n-1\right)+6\)

\(\Leftrightarrow n^3-3n^2+2n+12n-6n^2+6n-6=0\)

=>\(n^3-9n^2+20n-6=0\)

=>n=3

Số hạng nào hả bạn? 

tham khảo