## Cho khai triển biểu thức {\left( {{2^{\frac{{x – 1}}{2}}} + {2^{ – \frac{x}{3}}}} \right)^n} = C_n^0{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^n} + C_n

Question

Cho khai triển biểu thức
{\left( {{2^{\frac{{x – 1}}{2}}} + {2^{ – \frac{x}{3}}}} \right)^n} = C_n^0{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^n} + C_n^1{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^{n – 1}}\left( {{2^{ – \frac{x}{3}}}} \right) + … + C_n^n{\left( {{2^{ – \frac{x}{3}}}} \right)^n}
biết rằng trong khai triển đó C_n^3 = 5C_n^1 và số hạng thứ tư bằng 20n. Tìm n và x.

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3 tháng 2021-09-09T10:26:20+00:00 1 Answers 3 views 0

$$n=7$$ và $$x=4.$$
$$\begin{array}{l} {\left( {{2^{\frac{{x – 1}}{2}}} + {2^{ – \frac{x}{3}}}} \right)^n} = C_n^0{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^n} + C_n^1{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^{n – 1}}\left( {{2^{ – \frac{x}{3}}}} \right) + … + C_n^n{\left( {{2^{ – \frac{x}{3}}}} \right)^n}\\ C_n^3 = 5C_n^1 \Leftrightarrow \frac{{n!}}{{3!\left( {n – 3} \right)!}} = 5.\frac{{n!}}{{1!\left( {n – 1} \right)!}}\\ \Leftrightarrow \frac{1}{{6\left( {n – 3} \right)!}} = \frac{5}{{\left( {n – 1} \right)\left( {n – 2} \right)\left( {n – 3} \right)!}}\\ \Leftrightarrow \left( {n – 1} \right)\left( {n – 2} \right) = 30\\ \Leftrightarrow {n^2} – 3n + 2 – 30 = 0\\ \Leftrightarrow {n^2} – 3n – 28 = 0\\ \Leftrightarrow \left[ \begin{array}{l} n = – 4\,\,\left( {ktm} \right)\\ n = 7\,\,\,\left( {tm} \right) \end{array} \right..\\ So\,\,hang\,\,thu\,\,4\,\,bang\,\,20n\\ \Rightarrow C_n^3{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^{n – 3}}{\left( {{2^{ – \frac{x}{3}}}} \right)^3} = 20n\\ \Leftrightarrow C_7^3{\left( {{2^{\frac{{x – 1}}{2}}}} \right)^{7 – 3}}{\left( {{2^{ – \frac{x}{3}}}} \right)^3} = 20.7\\ \Leftrightarrow {35.2^{2\left( {x – 1} \right)}}{.2^{ – x}} = 140\\ \Leftrightarrow {2^{x – 2}} = 4\\ \Leftrightarrow x – 2 = 2\\ \Leftrightarrow x = 4. \end{array}$$