Câu hỏi của Vũ Thu Trang

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Giúo em câu 3, 5 ,6 8 của bài 1 với ạ. Cần gấp ạ

Nguyễn Việt Lâm · Accepted Answer

3.Cách 1: sử dụng đạo hàmXét khai triển: $\left(1+x\right)^{2010}=C_{2010}^0+xC_{2010}^1+x^2C_{2010}^2+...+x^nC_{2010}^{2010}$Đạo hàm 2 vế:$2010.\left(1+x\right)^{2009}=C_{2010}^1+2xC_{2010}^2+3x^2C_{2010}^3+...+2010.x^{2009}C_{2010}^{2010}$Cho $x=1$$\Rightarrow C_{2010}^1+2C_{2010}^2+...+2010.C_{2010}^{2010}=2010.2^{2009}$Cách 2:Ta có: $k.C_n^k=k.\dfrac{n!}{k!.\left(n-k\right)!}=n.\dfrac{\left(n-1\right)!}{\left(k-1\right)!.\left(n-k\right)!}=n.C_{n-1}^{k-1}$Do đó:$1.C_n^1=n.C_{n-1}^0$ ; $2.C_n^2=n.C_{n-1}^1$ ;....; $n.C_n^n=n.C_{n-1}^{n-1}$Cộng vế:$1.C_n^1+2.C_n^2+3.C_n^3+...+n.C_n^n=n\left(C_{n-1}^0+C_{n-1}^1+...+C_{n-1}^{n-1}\right)=n.2^{n-1}$Thay $n=2010$$\Rightarrow...$ (đpcm)Câu trả lời: 3.Cách 1: sử dụng đạo hàmXét khai triển: $\left(1+x\right)^{2010}=C_{2010}^0+xC_{2010}^1+x^2C_{2010}^2+...+x^nC_{2010}^{2010}$Đạo hàm 2 vế:$2010.\left(1+x\right)^{2009}=C_{2010}^1+2xC_{2010}^2+3x^2C_{2010}^3+...+2010.x^{2009}C_{2010}^{2010}$Cho $x=1$$\Rightarrow C_{2010}^1+2C_{2010}^2+...+2010.C_{2010}^{2010}=2010.2^{2009}$Cách 2:Ta có: $k.C_n^k=k.\dfrac{n!}{k!.\left(n-k\right)!}=n.\dfrac{\left(n-1\right)!}{\left(k-1\right)!.\left(n-k\right)!}=n.C_{n-1}^{k-1}$Do đó:$1.C_n^1=n.C_{n-1}^0$ ; $2.C_n^2=n.C_{n-1}^1$ ;....; $n.C_n^n=n.C_{n-1}^{n-1}$Cộng vế:$1.C_n^1+2.C_n^2+3.C_n^3+...+n.C_n^n=n\left(C_{n-1}^0+C_{n-1}^1+...+C_{n-1}^{n-1}\right)=n.2^{n-1}$Thay $n=2010$$\Rightarrow...$ (đpcm)

Nguyễn Việt Lâm · Answer

5.Áp dụng công thức: $C_n^k=C_n^{n-k}$$\Rightarrow C_{2011}^0=C_{2011}^{2011}$$C_{2011}^1=C_{2011}^{2010}$...$C_{2011}^{1005}=C_{2011}^{1006}$Cộng vế:$C_{2011}^0+C_{2011}^1+...+C_{2011}^{1005}=C_{2011}^{1006}+...+C_{2011}^{2010}+C_{2011}^{2011}$$\Rightarrow2\left(C_{2011}^0+C_{2011}^1+...+C_{2011}^{1005}\right)=C_{2011}^0+...+C_{2011}^{1005}+C_{2011}^{1006}+...++C_{2011}^{2011}$$\Rightarrow2\left(C_{2011}^0+C_{2011}^1+...+C_{2011}^{1005}\right)=2^{2011}$$\Rightarrow C_{2011}^0+C_{2011}^1+...+C_{2011}^{1005}=\dfrac{2^{2011}}{2}=2^{2010}$Câu trả lời: 5.Áp dụng công thức: $C_n^k=C_n^{n-k}$$\Rightarrow C_{2011}^0=C_{2011}^{2011}$$C_{2011}^1=C_{2011}^{2010}$...$C_{2011}^{1005}=C_{2011}^{1006}$Cộng vế:$C_{2011}^0+C_{2011}^1+...+C_{2011}^{1005}=C_{2011}^{1006}+...+C_{2011}^{2010}+C_{2011}^{2011}$$\Rightarrow2\left(C_{2011}^0+C_{2011}^1+...+C_{2011}^{1005}\right)=C_{2011}^0+...+C_{2011}^{1005}+C_{2011}^{1006}+...++C_{2011}^{2011}$$\Rightarrow2\left(C_{2011}^0+C_{2011}^1+...+C_{2011}^{1005}\right)=2^{2011}$$\Rightarrow C_{2011}^0+C_{2011}^1+...+C_{2011}^{1005}=\dfrac{2^{2011}}{2}=2^{2010}$

Nguyễn Việt Lâm · Answer

6.Áp dụng công thức tương tự bài 5:$C_{2n+1}^0=C_{2n+1}^{2n+1}$$C_{2n+1}^2=C_{2n+1}^{2n-1}$...$C_{2n+1}^{2n}=C_{2n+1}^1$Cộng vế:$\Rightarrow C_{2n+1}^0+C_{2n+1}^2+...+C_{2n+1}^n=C_{2n+1}^1+C_{2n+1}^3+...+C_{2n+1}^{2n+1}$ (1)$\Rightarrow2\left(C_{2n+1}^0+C_{2n+1}^2+...+C_{2n+1}^{2n}\right)=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}=2^{2n+1}$$\Rightarrow C_{2n+1}^0+C_{2n+1}^2+...+C_{2n+1}^{2n}=2^{2n}$ (2)Lại có:$C_{2n+1}^0=C_{2n+1}^{2n+1}$$C_{2n+1}^1=C_{2n+1}^{2n}$...$C_{2n+1}^n=C_{2n+1}^{n+1}$Cộng vế:$C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^n=C_{2n+1}^{n+1}+...+C_{2n+1}^{2n+1}$$\Rightarrow2\left(C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^n\right)=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}=2^{2n+1}$$\Rightarrow C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^n=2^{2n}$ (3)(1);(2);(3) suy ra đpcmCâu trả lời: 6.Áp dụng công thức tương tự bài 5:$C_{2n+1}^0=C_{2n+1}^{2n+1}$$C_{2n+1}^2=C_{2n+1}^{2n-1}$...$C_{2n+1}^{2n}=C_{2n+1}^1$Cộng vế:$\Rightarrow C_{2n+1}^0+C_{2n+1}^2+...+C_{2n+1}^n=C_{2n+1}^1+C_{2n+1}^3+...+C_{2n+1}^{2n+1}$ (1)$\Rightarrow2\left(C_{2n+1}^0+C_{2n+1}^2+...+C_{2n+1}^{2n}\right)=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}=2^{2n+1}$$\Rightarrow C_{2n+1}^0+C_{2n+1}^2+...+C_{2n+1}^{2n}=2^{2n}$ (2)Lại có:$C_{2n+1}^0=C_{2n+1}^{2n+1}$$C_{2n+1}^1=C_{2n+1}^{2n}$...$C_{2n+1}^n=C_{2n+1}^{n+1}$Cộng vế:$C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^n=C_{2n+1}^{n+1}+...+C_{2n+1}^{2n+1}$$\Rightarrow2\left(C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^n\right)=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}=2^{2n+1}$$\Rightarrow C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^n=2^{2n}$ (3)(1);(2);(3) suy ra đpcm

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