Bài 3: Nhị thức Niu-tơn

\(\left(a+b\right)^n=a^n+C^1_na^{n-1}b+C^2_na^{n-2}b^2+...+C^{n-1}_nab^{n-1}+b^n\)

=> \(\left(\sqrt{3}+\sqrt[3]{30}\right)^6=\sqrt{3}^6+C^1_6\sqrt{3}^5\cdot\sqrt[3]{30}+C^2_6\sqrt{3}^4\cdot\sqrt[3]{30}^2+C_6^3\sqrt{3}^3\cdot\sqrt[3]{30}^3+C^4_6\sqrt{3}^2\cdot\sqrt[3]{30}^4+C^5_6\sqrt{3}\cdot\sqrt[3]{30}^5+\sqrt[3]{30}^6\)

...muộn rồi, tự làm nốt nhé :))...

ta có : \(\left(a+b\right)^n=C^0_na^n+C^1_na^{n-1}b+C^2_na^{n-2}b^2+...+C^k_na^{n-k}b^k+...+C^n_nb^n\)

ta có : \(\left(\sqrt{3}+\sqrt[3]{30}\right)^6\)

\(=C^0_6\left(\sqrt{3}\right)^6+C^1_6\left(\sqrt{3}\right)^5\left(\sqrt[3]{30}\right)+C^2_6\left(\sqrt{3}\right)^4\left(\sqrt[3]{30}\right)^2+C^3_6\left(\sqrt{3}\right)^3\left(\sqrt[3]{30}\right)^3+C^4_6\left(\sqrt{3}\right)^2\left(\sqrt[3]{30}\right)^4+C^5_6\left(\sqrt{3}\right)\left(\sqrt[3]{30}\right)^5+C^6_6\left(\sqrt[3]{30}\right)^6\)

mk mới đọc sơ thôi ; không biết giải đúng hay sai ; mong mấy bạn đừng nép đá nha

ta có : \(\left(4x-8\right)^{10}=C^0_{10}\left(4x\right)^{10}+C^1_{10}\left(4x\right)^9\left(-8\right)+C^2_{10}\left(4a\right)^8\left(-8\right)^2+C^3_{10}\left(4a\right)^7\left(-8\right)^3+C^4_{10}\left(4a\right)^6\left(-8\right)^4+C^5_{10}\left(4a\right)^5\left(-8\right)^5+C^6_{10}\left(4a\right)^4\left(-8\right)^6+C^7_{10}\left(4a\right)^3\left(-8\right)^7+C^8_{10}\left(4a\right)^2\left(-8\right)^8+C^9_{10}\left(4a\right)\left(-8\right)^9+C^{10}_{10}\left(-8\right)^{10}\)

\(B=1.2+2.3+3.4+...+99.100\)

\(3B=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)

\(3B=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)

\(3B=99.100.101\)

\(B=\dfrac{99.100.101}{3}=333300\)

ta có : \(Q=C^1_n+2\dfrac{C_n^2}{C_n^1}+...+k\dfrac{C^k_n}{C_n^{k-1}}+...+n\dfrac{C^n_n}{C_n^{n-1}}\)

\(\Leftrightarrow Q=\dfrac{n!}{1!\left(n-1\right)!}+2\dfrac{1!\left(n-1\right)!}{2!\left(n-2\right)!}+...+k\dfrac{\left(k-1\right)!\left(n-k+1\right)!}{k!\left(n-k\right)!}+...+\dfrac{n\left(n-1\right)!1!}{n!}\)

\(\Leftrightarrow Q=n+\dfrac{2\left(n-1\right)}{2}+...+\dfrac{k\left(n-k+1\right)}{k}+...+\dfrac{n}{n}\)

\(\Leftrightarrow Q=n+\left(n-1\right)+...+\left(n-k+1\right)+...+1\)

\(\Leftrightarrow Q=n^2-\left(1+\left(1+1\right)+\left(1+2\right)+...+\left(n-1\right)\right)\)

ta có : \(\dfrac{\left(n+3\right)!}{n!}=\dfrac{1.2.3...n.\left(n+1\right)\left(n+2\right)\left(n+3\right)}{1.2.3...n}\)

\(\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

ta có : \(\left(1+\dfrac{2x}{3}\right)^{10}=\sum\limits^{10}_{k=0}\left(\dfrac{2}{3}\right)^k.x^k\)

vì \(0< \dfrac{2}{3}< 1\) \(\Rightarrow\left(\dfrac{2}{3}\right)^{k-1}>\left(\dfrac{2}{3}\right)^k\)

mà vì \(K\in N\)

\(\Rightarrow\) hệ số lớn nhất trong khai triển \(\left(1+\dfrac{2x}{3}\right)^{10}\) là \(\left(\dfrac{2}{3}\right)^0=1\)