Cho khai triển nhị thức Newton \({{\left( 2-3x \right)}^{2x}}\), biết rằng n là số nguyên dương thỏa mãn \(C_{2n+1}^{1}+C_{2n+1}^{3}+C_{2n+1}^{5}+...+C_{2n+1}^{2n+1}=1024\). Tìm hệ...

* Hướng dẫn giải

Ta có: \({{\left( x+1 \right)}^{2n+1}}=\sum\limits_{k=0}^{2n+1}{C_{2n+1}^{k}{{x}^{k}}}\)

Khi \(x=1\) ta có: \({{2}^{2n+1}}=\sum\limits_{k=0}^{2n+1}{C_{2n+1}^{k}}=C_{2n+1}^{0}+C_{2n+1}^{1}+C_{2n+1}^{2}+C_{2n+1}^{3}+C_{2n+1}^{4}+C_{2n+1}^{5}+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}\,\,\,\,\left( 1 \right)\)

Khi \(x=-1\) ta có: \(0=\sum\limits_{k=0}^{2n+1}{C_{2n+1}^{k}{{\left( -1 \right)}^{k}}}=C_{2n+1}^{0}-C_{2n+1}^{1}+C_{2n+1}^{2}-C_{2n+1}^{3}+C_{2n+1}^{4}-C_{2n+1}^{5}+...+C_{2n+1}^{2n}-C_{2n+1}^{2n+1}\,\,\left( 2 \right)\)

\(\begin{align}  & \left( 1 \right)-\left( 2 \right)\Rightarrow {{2}^{2n+1}}=2\left( C_{2n+1}^{1}+C_{2n+1}^{3}+C_{2n+1}^{5}+...+C_{2n+1}^{2n+1} \right)=2.1024=2048 \\  & \Leftrightarrow 2n+1=11\Leftrightarrow 2n=10\Leftrightarrow n=5 \\  & \Rightarrow {{\left( 2-3x \right)}^{2n}}={{\left( 2-3x \right)}^{10}}=\sum\limits_{k=0}^{10}{C_{10}^{k}{{.2}^{10-k}}.{{\left( -3x \right)}^{k}}}=\sum\limits_{k=0}^{10}{C_{10}^{k}{{.2}^{10-k}}.{{\left( -3 \right)}^{k}}.{{x}^{k}}} \\ \end{align}\)

Để tìm hệ số của \({{x}^{7}}\) ta cho \(k=7\Rightarrow \) Hệ số của  \({{x}^{7}}\) là \(C_{10}^{7}{{.2}^{3}}.{{\left( -3 \right)}^{7}}=-2099520\)

Chọn A.