Công thức nguyên hàm và tích phân
: Tính các tổng sau:${{S}_{2}}=C_{n}^{1}+2C_{n}^{2}+…+nC_{n}^{n}$ C. $2n{{.2}^{n-1}}$ B. $n{{.2}^{n+1}}$ C. $2n{{.2}^{n+1}}$ D. $n{{.2}^{n-1}}$ Hướng dẫn Chọn D Ta có: $kC_{n}^{k}=k.\frac{n!}{k!(n-k)!}=\frac{n!}{(k-1)!\text{ }\!\![\!\!\text{ }(n-1)-(k-1)\text{ }\!\!]\!\!\text{ }!}$ $=n\frac{(n-1)!}{(k-1)!\text{ …
: Tính các tổng sau:${{S}_{3}}=2.1.C_{n}^{2}+3.2C_{n}^{3}+4.3C_{n}^{4}+…+n(n-1)C_{n}^{n}$. C. $n(n-1){{2}^{n-2}}$ B. $n(n+2){{2}^{n-2}}$ C. $n(n-1){{2}^{n-3}}$ D. $n(n-1){{2}^{n+2}}$ Hướng dẫn Chọn A Ta có $k(k-1)C_{n}^{k}=\frac{n!}{(k-2)!(n-k)!}=n(n-1)C_{n-2}^{k-2}$ $\Rightarrow {{S}_{3}}=n(n-1)\sum\limits_{k=2}^{n}{C_{n-2}^{k-2}}=n(n-1){{2}^{n-2}}$.
: 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 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{ }\!\
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