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Xét khai triển:
\(\left(1+2x\right)^{2n+1}=C_{2n+1}^0+C_{2n+1}^1.2x+C_{2n+1}^2\left(2x\right)^2+...+C_{2n+1}^{2n+1}\left(2x\right)^{2n+1}\)
Đạo hàm 2 vế:
\(2\left(2n+1\right)\left(1+2x\right)^{2n}=2C_{2n+1}^1+2^2C_{2n+1}^2x+...+\left(2n+1\right)2^{2n+1}C_{2n+1}^{2n+1}x^{2n}\)
\(\Leftrightarrow\left(2n+1\right)\left(1+2x\right)^{2n}=C_{2n+1}^1+2C_{2n+1}^2x+...+\left(2n+1\right)2^{2n}C_{2n+1}^{2n+1}x^{2n}\)
Cho \(x=-1\) ta được:
\(2n+1=C_{2n+1}^1-2C_{2n+1}^2+...+\left(2n+1\right)2^{2n}C_{2n+1}^{2n+1}\)
\(\Rightarrow2n+1=2019\Rightarrow n=1009\)
Ta có : \(C^k_{2n+1}=C^{2n+1-k}_{2n+1}\)
\(\Rightarrow2VT=C^1_{2n+1}+C^2_{2n+1}+...+C^{2n}_{2n+1}=2^{21}-2\)
\(\Leftrightarrow2^{2n+1}-C^0_{2n+1}-C^{2n+1}_{2n+1}=2^{21}-2\)
\(\Leftrightarrow2n+1=21\Leftrightarrow n=10\)
\(\sum\limits^{2n+1}_{k=0}C^k_{2n+1}=\left(1+1\right)^{2n+1}=2^{2n+1}\)
Lại có \(C^0_{2n+1}+C^1_{2n+1}+...+C^n_{2n+1}=C^{2n+1}_{2n+1}+C^{2n}_{2n+1}+...+C^{n+1}_{2n+1}\)
\(\Rightarrow C^0_{2n+1}+C^1_{2n+1}+...C^n_{2n+1}=\dfrac{2^{2n+1}}{2}\)
\(\Leftrightarrow2^{20}-1=2^{2n}-C^0_{2n+1}\)
\(\Leftrightarrow2^{20}-1=2^{2n}-1\)
\(\Leftrightarrow2n=20\)
\(\Leftrightarrow n=10\)
Xét khai triển:
\(\left(x+1\right)^{2n+1}=C_{2n+1}^0+C_{2n+1}^1x+C_{2n+1}^2x^2+...+C_{2n+1}^{2n+1}x^{2n+1}\)
Cho \(x=1\) ta được:
\(2^{2n+1}=C_{2n+1}^0+C_{2n+1}^1+C_{2n+1}^2+...+C_{2n+1}^{2n+1}\)
\(=1+C_{2n+1}^1+C_{2n+1}^2+...+C_{2n+1}^n+C_{2n+1}^{n+1}+...+C_{2n+1}^{2n}+1\)
\(=1+C_{2n+1}^1+...+C_{2n+1}^n+C_{2n+1}^n+...+C_{2n+1}^1+1\)
\(=2\left(1+C_{2n+1}^1+C_{2n+1}^2+...+C_{2n+1}^n\right)\)
\(\Rightarrow2^{2n}-1=C_{2n+1}^1+C_{2n+1}^2+...+C_{2n+1}^n\)
\(\Rightarrow2^{2n-1}=2^{20}-1\Rightarrow2n=20\Rightarrow n=10\)
Khai triển: \(\left(x^2-x-1\right)^{10}\)
\(\left\{{}\begin{matrix}k_0+k_1+k_2=10\\k_1+2k_2=6\end{matrix}\right.\) \(\Rightarrow\left(k_0;k_1;k_2\right)=\left(4;6;0\right);\left(5;4;1\right);\left(6;2;2\right);\left(7;0;3\right)\)
Hệ số của \(x^6:\)
\(\frac{10!}{4!.6!}+\frac{10!}{5!.4!}.\left(-1\right)^5+\frac{10!}{6!.2!.2!}+\frac{10!}{7!.3!}.\left(-1\right)^7\)
Giả thiết tương đương:
\(C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}=2^{100}\) (thay \(1=C_{2n+1}^{2n+1}\))
Mặt khác:
\(C_{2n+1}^{2n+1}=C_{2n+1}^0\)
\(C_{2n+1}^{2n}=C_{2n+1}^1\)
....
\(C_{2n+1}^{n+1}=C_{2n+1}^n\)
Cộng vế:
\(\Rightarrow C_{2n+1}^{n+1}+C_{2n+1}^{n+2}+...+C_{2n+1}^{2n+1}=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^n\)
\(\Rightarrow2\left(C_{2n+1}^{n+1}+...+C_{2n+1}^{2n+1}\right)=C_{2n+1}^0+C_{2n+1}^1+...+C_{2n+1}^{2n+1}\)
\(\Rightarrow2.2^{100}=2^{2n+1}\) (đẳng thức cơ bản: \(\sum\limits^n_{k=0}C_n^k=2^n\))
\(\Leftrightarrow2^{101}=2^{2n+1}\)
\(\Rightarrow2n+1=101\)
\(\Rightarrow n=50\)
SHTQ trong khai triển: \(C_{50}^k.\left(x^{-3}\right)^k.\left(x^2\right)^{50-k}=C_{50}^kx^{100-5k}\)
\(100-5k=20\Rightarrow k=16\)
Hệ số: \(C_{50}^{16}\)
Xài cái này gõ bài đi bạn, thề như này hiểu chết liền á :(
Xét khai triển
\(\left(x+1\right)^{2n+1}=C_{2n+1}^0+C_{2n+1}^1x+...+C_{2n+1}^{2n}x^{2n}+C_{2n+1}^{2n+1}x^{2n+1}\)
Cho \(x=1\) ta được:
\(2^{2n+1}=C^0_{2n+1}+C_{2n+1}^1+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}\)
\(\Leftrightarrow2^{2n+1}=2+C_{2n+1}^1+C_{2n+1}^2+...+C_{2n+1}^{2n}\)
\(\Leftrightarrow2^{2n+1}-2=C_{2n+1}^1+C_{2n+1}^2+...+C_{2n+1}^{2n}\)
\(\Leftrightarrow2^{10}-1=2^{2n+1}-2\Rightarrow2^{2n+1}=2^{10}+1\)
Không tồn tại n thỏa mãn yêu cầu bài toán (bạn xem lại đề bài)