Tháng Năm 4, 2024

: Tính tổng sau: ${{S}_{1}}={{5}^{n}}C_{n}^{0}+{{5}^{n-1}}.3.C_{n}^{n-1}+{{3}^{2}}{{.5}^{n-2}}C_{n}^{n-2}+…+{{3}^{n}}C_{n}^{0}$

by adminTháng Chín 27, 2021

: Tính tổng sau: ${{S}_{1}}={{5}^{n}}C_{n}^{0}+{{5}^{n-1}}.3.C_{n}^{n-1}+{{3}^{2}}{{.5}^{n-2}}C_{n}^{n-2}+…+{{3}^{n}}C_{n}^{0}$

C. ${{28}^{n}}$

B. $1+{{8}^{n}}$

C. ${{8}^{n-1}}$

D. ${{8}^{n}}$

Hướng dẫn

Chọn D

Ta có: ${{S}_{1}}={{(5+3)}^{n}}={{8}^{n}}$

admin

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Đưa thừa số vào trong dấu căn \(\begin{array}{l}a)\,\,\frac{1}{{xy}}\sqrt {\frac{{{x^2}{y^2}}}{2}} & & & b)\,\,a\sqrt 2 & & & c)\,\, – \frac{a}{b}\sqrt {\frac{b}{a}} \,\,\,\left( {a > 0,\,\,\,b > 0} \right)\\d)\,\,a\sqrt {\frac{3}{a}} & & & e)\,\,\frac{1}{{2x – 1}}\sqrt {5\left( {1 – 4x + 4{x^2}} \right)} .\end{array}\) A \(\begin{array}{l} a)\,\,\left\{ \begin{array}{l} \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\ – \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0 \end{array} \right.\\ b)\,\,\left\{ \begin{array}{l} \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ – \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0 \end{array} \right.\\ c)\,\,\sqrt {\frac{a}{b}} \\ d)\,\,\left\{ \begin{array}{l} \sqrt 5 \,\,\,khi\,\,\,x > \frac{1}{2}\\ – \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2} \end{array} \right. \end{array}\) B \(\begin{array}{l} a)\,\,\left\{ \begin{array}{l} \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\ – \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0 \end{array} \right.\\ b)\,\,\left\{ \begin{array}{l} \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ – \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0 \end{array} \right.\\ c)\,\, – \sqrt {\frac{a}{b}} \\ d)\,\,\left\{ \begin{array}{l} \sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\ – \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2} \end{array} \right. \end{array}\) C \(\begin{array}{l} a)\,\,\frac{{\sqrt 2 }}{2}\\ b)\,\,\left\{ \begin{array}{l} \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ – \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0 \end{array} \right.\\ c)\,\,\sqrt {\frac{a}{b}} \\ d)\,\,\left\{ \begin{array}{l} \sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\ – \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2} \end{array} \right. \end{array}\) D \(\begin{array}{l} a)\,\,\frac{{\sqrt 2 }}{2}\\ b)\,\,\left\{ \begin{array}{l} \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ – \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0 \end{array} \right.\\ c)\,\,\sqrt {\frac{a}{b}} \\ d)\,\,\sqrt 5 \end{array}\)