\(C_n^0+C_n^1+C_n^2=11\)

\(\Rightarrow1+n+\dfrac{n\left(n-1\right)}{2}=11\)

\(\Leftrightarrow n^2+n-20=0\Rightarrow\left[{}\begin{matrix}n=4\\n=-5\left(loại\right)\end{matrix}\right.\)

\(\left(x^3+\dfrac{1}{x^2}\right)^4\) có SHTQ: \(C_4^k.x^{3k}.x^{-2\left(4-k\right)}=C_4^k.x^{5k-8}\)

\(5k-8=7\Rightarrow k=3\)

Hệ số: \(C_4^3=4\)

\(Bài.1:u_n=\dfrac{3}{2}.\left(\dfrac{1}{2}\right)^n=\dfrac{3}{512}\\ \Rightarrow\left(\dfrac{1}{2}\right)^n=\dfrac{3}{512}:\dfrac{3}{2}=\dfrac{1}{256}=\dfrac{1}{2^8}\\ Mà:\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^8\\ Vậy:n=8\\ \Rightarrow Vậy:\dfrac{3}{512}.là.số.hạng.thứ.8\)

\(x_{n+1}=\dfrac{1}{2}x_n+2^{n-2}\Leftrightarrow x_{n+1}-\dfrac{1}{6}.2^{n+1}=\dfrac{1}{2}\left(x_n-\dfrac{1}{6}.2^n\right)\)

Đặt \(x_n-\dfrac{1}{6}.2^n=y_n\Rightarrow\left\{{}\begin{matrix}y_1=x_1-\dfrac{1}{6}.2^1=\dfrac{8}{3}\\y_{n+1}=\dfrac{1}{2}y_n\end{matrix}\right.\)

\(\Rightarrow y_n\) là CSN với công bội \(q=\dfrac{1}{2}\)

\(\Rightarrow y_n=\dfrac{8}{3}.\left(\dfrac{1}{2}\right)^{n-1}=\dfrac{4}{3.2^n}\)

\(\Rightarrow x_n=y_n+\dfrac{1}{6}.2^n=\dfrac{4}{3.2^n}+\dfrac{2^n}{6}\)

Câu 2:

\(\Leftrightarrow\dfrac{\left(n+2\right)!}{2!\cdot n!}-4\cdot\dfrac{\left(n+1\right)!}{n!\cdot1!}=2\left(n+1\right)\)

\(\Leftrightarrow\dfrac{\left(n+1\right)\left(n+2\right)}{2}-4\cdot\dfrac{n+1}{1}=2\left(n+1\right)\)

\(\Leftrightarrow\left(n+1\right)\left(n+2\right)-8\left(n+1\right)=4\left(n+1\right)\)

=>(n+1)(n+2-8-4)=0

=>n=-1(loại) hoặc n=10

=>\(A=\left(\dfrac{1}{x^4}+x^7\right)^{10}\)

SHTQ là: \(C^k_{10}\cdot\left(\dfrac{1}{x^4}\right)^{10-k}\cdot x^{7k}=C^k_{10}\cdot1\cdot x^{11k-40}\)

Số hạng chứa x^26 tương ứng với 11k-40=26

=>k=6

=>Số hạng cần tìm là: \(210x^{26}\)

\(\left(x^{-\frac{2}{3}}+x^{\frac{3}{4}}\right)^{17}=\sum\limits^{17}_{k=0}C_{17}^k\left(x^{-\frac{2}{3}}\right)^k\left(x^{\frac{3}{4}}\right)^{17-k}=\sum\limits^{17}_{k=0}C_{17}^kx^{\frac{51}{4}-\frac{17}{12}k}\)

Số hạng thứ 13 \(\Rightarrow k=12\) là: \(C_{17}^{12}x^{-\frac{17}{4}}\)

b/ Xét khai triển:

\(\left(3-x\right)^n=C_n^03^n+C_n^13^{n-1}\left(-x\right)^1+C_n^23^{n-2}\left(-x\right)^2+...+C_n^n\left(-x\right)^n\)

Cho \(x=1\) ta được:

\(2^n=3^nC_n^0-3^{n-1}C_n^1+3^{n-2}C_n^2+...+\left(-1\right)^nC_n^n\)

À, đến đây mới thấy đề thiếu, biết rằng cái kia làm sao hả bạn?

\(\left(x+x^{-1}\right)^n=\sum\limits^n_{k=0}C_n^kx^k\left(x^{-1}\right)^{n-k}=\sum\limits^n_{k=0}C_n^kx^{2k-n}\)

Theo bài ra ta có: \(C_n^2-C_n^1=35\)

\(\Leftrightarrow\frac{n!}{2!\left(n-2\right)!}-\frac{n!}{\left(n-1\right)!}=35\)

\(\Leftrightarrow\frac{n\left(n-1\right)}{2}-n=35\)

\(\Leftrightarrow n^2-3n-70=0\Rightarrow n=10\)

Số hạng ko chứa x \(\Rightarrow2k-n=0\Rightarrow k=\frac{n}{2}=5\)

Số hạng đó là \(C_{10}^5\)

Câu 2:

\(\left\{{}\begin{matrix}u_1+u_5-u_3=10\\u_1+u_6=17\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u_1+u_1+4d-u_1-2d=10\\u_1+u_1+5d=17\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u_1+2d=10\\2u_1+5d=17\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2u_1+4d=20\\2u_1+5d=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2u_1+4d-2u_1-5d=20-17\\2u_1+5d=17\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-d=3\\2u_1+5d=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}d=-3\\2u_1=17-5d=17+5\cdot3=32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u_1=16\\d=-3\end{matrix}\right.\)

Câu 1:

Để a,b,c lập thành cấp số cộng thì

\(\left[{}\begin{matrix}a+c=2b\\a+b=2c\\b+c=2a\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x+1+x^2-1=2\cdot\left(3x-2\right)\\x+1+3x-2=2\left(x^2-1\right)\\x^2-1+3x-2=2\left(x+1\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x^2+x-6x+4=0\\2x^2-2=4x-1\\x^2+3x-3-2x-2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x^2-5x+4=0\\2x^2-4x-1=0\\x^2+x-5=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\left(x-1\right)\left(x-4\right)=0\\2x^2-4x-1=0\\x^2+x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\left\{1;4\right\}\\x\in\left\{\dfrac{2+\sqrt{6}}{2};\dfrac{2-\sqrt{6}}{2}\right\}\\x\in\left\{\dfrac{-1+\sqrt{21}}{2};\dfrac{-1-\sqrt{21}}{2}\right\}\end{matrix}\right.\)