\(C_2^2+C_3^2+...+C_n^2=C_3^3+C_3^2+C_4^2+...+C_n^2\) (do \(C_2^2=C_3^3=1\))

\(=C_4^3+C_4^2+C_5^2+...+C_n^2=C_5^3+C_5^2+...+C_n^2\)

\(=...=C_n^3+C_n^2=C_{n+1}^3\)

Do đó:

\(2C_{n+1}^3=3A_{n+1}^2\Leftrightarrow\dfrac{2.\left(n+1\right)!}{3!.\left(n-2\right)!}=\dfrac{3.\left(n+1\right)!}{\left(n-1\right)!}\)

\(\Leftrightarrow n-1=9\Rightarrow n=10\)

\(\Rightarrow P=\left(1-x-3x^3\right)^{10}=\sum\limits^{10}_{k=0}C_{10}^k\left(-x-3x^3\right)^k\)

\(=\sum\limits^{10}_{k=0}C_{10}^k\left(-1\right)^k\left(x+3x^3\right)^k=\sum\limits^{10}_{k=0}\sum\limits^k_{i=0}C_{10}^kC_k^i\left(-1\right)^kx^i.3^{k-i}.x^{3\left(k-i\right)}\)

\(=\sum\limits^{10}_{k=0}\sum\limits^k_{i=0}C_{10}^kC_k^i\left(-1\right)^k.3^{k-i}.x^{3k-2i}\)

Ta có: \(\left\{{}\begin{matrix}0\le i\le k\le10\\i;k\in N\\3k-2i=4\end{matrix}\right.\) \(\Rightarrow\left(i;k\right)=\left(1;2\right);\left(4;4\right)\)

Hệ số: \(C_{10}^2C_2^1\left(-1\right)^2.3^1+C_{10}^4C_4^4.\left(-1\right)^4.3^0=...\)

\(\Rightarrow he-so:\left[{}\begin{matrix}C^9_{10}C^1_9\left(-3\right)^{10-9}\left(-1\right)=270\\C^{10}_{10}C^4_{10}\left(-3\right)^{10-10}.\left(-1\right)^4=210\end{matrix}\right.\)

ta có : \(\left(1-3x\right)^n=\sum\limits^n_{k=0}C^k_n\left(1\right)^{n-k}\left(-3\right)^k\left(x^k\right)\)

để có \(x^2\) trong khai triển thì \(k=2\)

khi đó hệ số của số hạng chứa \(x^2\) là \(\)\(C^2_n\left(-3\right)^2=90\)

\(\Leftrightarrow C^2_n=10\Leftrightarrow\dfrac{n!}{2!\left(n-2\right)!}=10\) \(\Leftrightarrow\dfrac{n\left(n-1\right)}{2}=10\)

\(\Leftrightarrow n^2-n-20=0\left[{}\begin{matrix}n=5\left(N\right)\\n=-4\left(L\right)\end{matrix}\right.\) vậy \(n=5\)

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

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

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

\(\left(x^3+\dfrac{1}{x^2}\right)^4\) 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: \(C_4^3=4\)

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: \(C_5^3.2^2\)

\(\left(3-1\right)^n=1024\Leftrightarrow2^n=2^{10}\Rightarrow n=10\)

\(\left(3-x^2\right)^{10}\) có SHTQ: \(C_{10}^k.3^k.\left(-1\right)^{10-k}.x^{20-2k}\)

Số hạng chứa \(x^{12}\Rightarrow20-2k=12\Rightarrow k=4\)

Hệ số: \(C_{10}^4.3^4=...\)

Đáp án A.