chứng minh \(1-C^1_n+C^2_n-C^3_n+..+\left(-1\right)^kC^k_n=\left(-1\right)^kC^k_{n-1}\)
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Phạm Dương Ngọc Nhi thế thì bạn học pp này đi. Cái pp này giúp cm nhiều bài một cách dễ dàng
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét khai triển:
\(\left(1+x\right)^n=C_n^0+C_n^1x+C_n^2x^2+...+C_n^nx^n\)
\(\Leftrightarrow x\left(1+x\right)^n=C_n^0x+C_n^1x^2+C_n^2x^3+...+C_n^nx^{n+1}\)
Đạo hàm 2 vế:
\(\left(1+x\right)^n+nx\left(1+x\right)^{n-1}=C_n^0+2C_n^1x+3C_n^2x^2+...+\left(n+1\right)C_n^nx^n\)
Thay \(x=1\)
\(\Rightarrow2^n+n.2^{n-1}=1+2C_n^1+3C_n^2+...+\left(n+1\right)C_n^n\)
\(\Rightarrow2^{n-1}\left(2+n\right)-1=111\)
\(\Rightarrow2^{n-1}\left(2+n\right)=112=2^4.7\)
\(\Rightarrow n=5\)
\(\left(x^2+\dfrac{2}{x}\right)^5=\sum\limits^5_{k=0}C_5^kx^{2k}.2^{5-k}.x^{k-5}=\sum\limits^5_{k=0}C_5^k.2^{5-k}.x^{3k-5}\)
\(3k-5=4\Rightarrow k=3\Rightarrow\) hệ số: \(C_5^3.2^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 )
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(1C_n^1+2C_n^2+...+nC_n^n\)
\(=n.C_{n-1}^0+nC_{n-1}^1+...+n\left(C_{n-1}^{n-1}\right)\)
\(=n\left(C_{n-1}^0+C_{n-1}^1+...+C_{n-1}^{n-1}\right)\)
\(=n.2^{n-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\frac{n+1}{n+2}\left(\frac{1}{C^k_{n+1}}+\frac{1}{C^{k+1}_{n+1}}\right)=\frac{n+1}{n+2}.\frac{k!\left(n+1-k\right)!+\left(k+1\right)!\left(n-k\right)!}{\left(n+1\right)!}\)
\(=\frac{1}{n+2}.\frac{k!\left(n-k\right)!}{n!}\left[\left(n+1-k\right)+\left(k+1\right)\right]=\frac{k!\left(n-k\right)!}{n!}=\frac{1}{C^k_n}=VP\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(n=8\)