Thẻ: Toán lớp 11

: Tính tổng $S=C_{n}^{0}+\frac{{{3}^{2}}-1}{2}C_{n}^{1}+…+\frac{{{3}^{n+1}}-1}{n+1}C_{n}^{n}$ C. $S=\frac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}$ B. $S=\frac{{{4}^{n+1}}+{{2}^{n+1}}}{n+1}-1$ C. $S=\frac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}+1$ D. $S=\frac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}-1$ Hướng dẫn Chọn D Ta có $S={{S}_{1}}-{{S}_{2}}$, trong đó ${{S}_{1}}=C_{n}^{0}+\frac{{{3}^{2}}}{2}C_{n}^{1}+\frac{{{3}^{3}}}{3}C_{n}^{2}+…+\frac{{{3}^{n+1}}}{n+1}C_{n}^{n}$ ${{S}_{2}}=\frac{1}{2}C_{n}^{1}+\frac{1}{3}C_{n}^{2}+…+\frac{1}{n+1}C_{n}^{n}$ Ta có …

: Tính tổng $S=C_{n}^{0}+\frac{{{2}^{2}}-1}{2}C_{n}^{1}+…+\frac{{{2}^{n+1}}-1}{n+1}C_{n}^{n}$ C. $S=\frac{{{3}^{n+1}}-{{2}^{n+1}}}{n+1}$ B. $S=\frac{{{3}^{n}}-{{2}^{n+1}}}{n+1}$ C. $S=\frac{{{3}^{n+1}}-{{2}^{n}}}{n+1}$ D. $S=\frac{{{3}^{n+1}}+{{2}^{n+1}}}{n+1}$ Hướng dẫn Chọn A Ta có: $S={{S}_{1}}-{{S}_{2}}$ Trong đó ${{S}_{1}}=\sum\limits_{k=0}^{n}{C_{n}^{k}\frac{{{2}^{k+1}}}{k+1}};\text{ }{{S}_{2}}=\sum\limits_{k=0}^{n}{\frac{C_{n}^{k}}{k+1}}=\frac{{{2}^{n+1}}-1}{n+1}-1$ Mà $\frac{{{2}^{k+1}}}{k+1}C_{n}^{k}=\frac{{{2}^{k+1}}}{n+1}C_{n+1}^{k+1}$$\Rightarrow …

: Tìm số nguyên dương n sao cho : $C_{2n+1}^{1}-2.2C_{2n+1}^{2}+{{3.2}^{2}}C_{2n+1}^{3}-…+(2n+1){{2}^{n}}C_{2n+1}^{2n+1}=2005$ C. n=1001 B. n=1002 C. n=1114 D. n=102 Hướng dẫn Chọn B Đặt $S=\sum\limits_{k=1}^{2n+1}{{{(-1)}^{k-1}}.k{{.2}^{k-1}}C_{2n+1}^{k}}$ Ta …

: Tính tổng${{1.3}^{0}}{{.5}^{n-1}}C_{n}^{n-1}+{{2.3}^{1}}{{.5}^{n-2}}C_{n}^{n-2}+…+n{{.3}^{n-1}}{{5}^{0}}C_{n}^{0}$ C. $n{{.8}^{n-1}}$ B. $(n+1){{.8}^{n-1}}$C.$(n-1){{.8}^{n}}$ D. $n{{.8}^{n}}$ Hướng dẫn Chọn A Ta có: $VT=\sum\limits_{k=1}^{n}{k{{.3}^{k-1}}{{.5}^{n-k}}C_{n}^{n-k}}$ Mà $k{{.3}^{k-1}}{{.5}^{n-k}}C_{n}^{n-k}=n{{.3}^{k-1}}{{.5}^{n-k}}.C_{n-1}^{k-1}$ Suy ra: $VT=n({{3}^{0}}{{.5}^{n-1}}C_{n-1}^{0}+{{3}^{1}}{{.5}^{n-2}}C_{n-1}^{1}+…+{{3}^{n-1}}{{5}^{0}}C_{n-1}^{n-1})$ $=n{{(5+3)}^{n-1}}=n{{.8}^{n-1}}$

: Tính tổng $S=2.1C_{n}^{2}+3.2C_{n}^{3}+4.3C_{n}^{4}+…+n(n-1)C_{n}^{n}$ C. $n(n+1){{2}^{n-2}}$ B. $n(n-1){{2}^{n-2}}$ C. $n(n-1){{2}^{n}}$ D. $(n-1){{2}^{n-2}}$ Hướng dẫn Chọn B Ta có: $S=\sum\limits_{k=2}^{n}{k(k-1)C_{n}^{k}}$ Mà $k(k-1)C_{n}^{k}=n(n-1)C_{n-2}^{k-2}$ Suy ra $S=n(n-1)(C_{n-2}^{0}+C_{n-2}^{1}+C_{n-2}^{2}+…+C_{n-2}^{n-2})=n(n-1){{2}^{n-2}}$

: Tính tổng ${{\left( C_{n}^{0} \right)}^{2}}+{{\left( C_{n}^{1} \right)}^{2}}+{{\left( C_{n}^{2} \right)}^{2}}+…+{{\left( C_{n}^{n} \right)}^{2}}$ C. $C_{2n}^{n}$ B. $C_{2n}^{n-1}$ C. $2C_{2n}^{n}$ D. $C_{2n-1}^{n-1}$ Hướng dẫn Chọn A Ta …

: Tính các tổng sau: ${{S}_{1}}=C_{n}^{0}+\frac{1}{2}C_{n}^{1}+\frac{1}{3}C_{n}^{2}+…+\frac{1}{n+1}C_{n}^{n}$ C. $\frac{{{2}^{n+1}}+1}{n+1}$ B. $\frac{{{2}^{n+1}}-1}{n+1}$ C. $\frac{{{2}^{n+1}}-1}{n+1}+1$ D. $\frac{{{2}^{n+1}}-1}{n+1}-1$ Hướng dẫn Chọn B Ta có: $\frac{1}{k+1}C_{n}^{k}=\frac{1}{k+1}\frac{n!}{k!(n-k)!}=\frac{1}{n+1}\frac{(n+1)!}{(k+1)!\text{ }\!\![\!\!\text{ (}n+1)-(k+1))!}$ $=\frac{1}{n+1}C_{n+1}^{k+1}$ (*) …

: Tìm số nguyên dương n sao cho: $C_{n}^{0}+2C_{n}^{1}+4C_{n}^{2}+…+{{2}^{n}}C_{n}^{n}=243$ C. 4 B. 11 C. 12 D. 5 Hướng dẫn Chọn D Xét khai triển: ${{(1+x)}^{n}}=C_{n}^{0}+xC_{n}^{1}+{{x}^{2}}C_{n}^{2}+…+{{x}^{n}}C_{n}^{n}$ …

: Khai triển ${{\left( 1+x+{{x}^{2}}+{{x}^{3}} \right)}^{5}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+…+{{a}_{15}}{{x}^{15}}$ a) Hãy tính hệ số ${{a}_{10}}$. C. ${{a}_{10}}=C_{5}^{0}.+C_{5}^{4}+C_{5}^{4}C_{5}^{3}$ B. ${{a}_{10}}=C_{5}^{0}.C_{5}^{5}+C_{5}^{2}C_{5}^{4}+C_{5}^{4}C_{5}^{3}$ C. ${{a}_{10}}=C_{5}^{0}.C_{5}^{5}+C_{5}^{2}C_{5}^{4}-C_{5}^{4}C_{5}^{3}$ D. ${{a}_{10}}=C_{5}^{0}.C_{5}^{5}-C_{5}^{2}C_{5}^{4}+C_{5}^{4}C_{5}^{3}$ b) Tính tổng $T={{a}_{0}}+{{a}_{1}}+…+{{a}_{15}}$ và …

: Khai triển ${{\left( 1+2x+3{{x}^{2}} \right)}^{10}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+…+{{a}_{20}}{{x}^{20}}$ a) Hãy tính hệ số ${{a}_{4}}$ C. ${{a}_{4}}=C_{10}^{0}{{.2}^{4}}$ B. ${{a}_{4}}={{2}^{4}}C_{10}^{4}$ C. ${{a}_{4}}=C_{10}^{0}C_{10}^{4}$ D. ${{a}_{4}}=C_{10}^{0}{{.2}^{4}}C_{10}^{4}$ b) Tính tổng $S={{a}_{1}}+2{{a}_{2}}+4{{a}_{3}}+…+{{2}^{20}}{{a}_{20}}$ C. …