Dạng bài tính tổng trong khai triển nhị thức Newton, Đại số giải tích 11

Chọn A

Tính chất của khai triển nhị thức Niu – Tơn.

Chọn A

$\text{S =}\,\,\text{C}_{\text{6}}^{\text{0}}\text{+C}_{\text{6}}^{\text{1}}\text{+}…\text{+C}_{\text{6}}^{\text{6}}={{2}^{6}}=64$

Chọn A

Với $x=1,y=1$ ta có $\text{S=}\,\,\text{C}_{\text{5}}^{\text{0}}\text{+C}_{\text{5}}^{\text{1}}\text{+}…\text{+C}_{\text{5}}^{\text{5}}={{(1+1)}^{5}}=32$.

Chọn D

Xét khai triển: ${{(1+x)}^{n}}=C_{n}^{0}+xC_{n}^{1}+{{x}^{2}}C_{n}^{2}+…+{{x}^{n}}C_{n}^{n}$

Cho $x=2$ ta có: $C_{n}^{0}+2C_{n}^{1}+4C_{n}^{2}+…+{{2}^{n}}C_{n}^{n}={{3}^{n}}$

Do vậy ta suy ra ${{3}^{n}}=243={{3}^{5}}\Rightarrow n=5$.

Chọn A

Với $x=1,y=1$ ta có $\text{S=}\,\,\text{C}_{\text{5}}^{\text{0}}\text{+C}_{\text{5}}^{\text{1}}\text{+}…\text{+C}_{\text{5}}^{\text{5}}={{(1+1)}^{5}}=32$.

Đặt $f(x)={{(1+x+{{x}^{2}}+{{x}^{3}})}^{5}}={{(1+x)}^{5}}{{(1+{{x}^{2}})}^{5}}$

a) Do đó hệ số ${{x}^{10}}$bằng: ${{a}_{10}}=C_{5}^{0}.C_{5}^{5}+C_{5}^{2}C_{5}^{4}+C_{5}^{4}C_{5}^{3}$

b) $T=f(1)={{4}^{5}}$; $S=f(-1)=0$

Đặt $f(x)={{(1+2x+3{{x}^{2}})}^{10}}=\sum\limits_{k=0}^{10}{C_{10}^{k}{{3}^{k}}{{x}^{2k}}{{(1+2x)}^{10-k}}}$

$=\sum\limits_{k=0}^{10}{C_{10}^{k}{{3}^{k}}{{x}^{2k}}\sum\limits_{i=0}^{10-k}{C_{10-k}^{i}{{2}^{10-k-i}}{{x}^{10-k-i}}}}$

$=\sum\limits_{k=0}^{10}{\sum\limits_{i=0}^{10-k}{C_{10}^{k}C_{10-k}^{i}{{3}^{k}}{{2}^{10-k-i}}{{x}^{10+k-i}}}}$

a) Ta có: ${{a}_{4}}=C_{10}^{0}{{.2}^{4}}C_{10}^{4}+$

b) Ta có $S=f(2)={{17}^{10}}$

Chọn A

Ta có: $S=\dfrac{1}{2}\left( C_{n}^{0}-\dfrac{1}{2}C_{n}^{1}+\dfrac{1}{3}C_{n}^{2}-…+\dfrac{{{(-1)}^{n}}}{n+1}C_{n}^{n} \right)$

Vì $\dfrac{{{(-1)}^{k}}}{k+1}C_{n}^{k}=\dfrac{{{(-1)}^{k}}}{n+1}C_{n+1}^{k+1}$ nên:$S=\dfrac{1}{2(n+1)}\sum\limits_{k=0}^{n}{{{(-1)}^{k}}C_{n+1}^{k+1}}$

$=\dfrac{-1}{2(n+1)}\left( \sum\limits_{k=0}^{n+1}{{{(-1)}^{k}}C_{n+1}^{k}}-C_{n+1}^{0} \right)=\dfrac{1}{2(n+1)}$.

Chọn A

Ta có: $S={{3}^{n}}\sum\limits_{k=1}^{n}{kC_{n}^{k}{{\left( \dfrac{1}{3} \right)}^{k}}}$

Vì $kC_{n}^{k}{{\left( \dfrac{1}{3} \right)}^{k}}=n{{\left( \dfrac{1}{3} \right)}^{k}}C_{n-1}^{k-1}$ $\forall k\ge 1$nên

$S={{3}^{n}}.n\sum\limits_{k=1}^{n}{{{\left( \dfrac{1}{3} \right)}^{k}}C_{n-1}^{k-1}}={{3}^{n-1}}.n\sum\limits_{k=0}^{n-1}{{{\left( \dfrac{1}{3} \right)}^{k}}C_{n-1}^{k}}$$={{3}^{n-1}}.n{{(1+\dfrac{1}{3})}^{n-1}}=n{{.4}^{n-1}}$.

Chọn B

Ta có:

$\dfrac{1}{k+1}C_{n}^{k}=\dfrac{1}{k+1}\dfrac{n!}{k!(n-k)!}=\dfrac{1}{n+1}\dfrac{(n+1)!}{(k+1)!\text{ }\!\![\!\!\text{ (}n+1)-(k+1))!}$

$=\dfrac{1}{n+1}C_{n+1}^{k+1}$ (*)

$\Rightarrow {{S}_{1}}=\dfrac{1}{n+1}\sum\limits_{k=0}^{n}{C_{n+1}^{k+1}}=\dfrac{1}{n+1}\left( \sum\limits_{k=0}^{n+1}{C_{n+1}^{k}}-C_{n+1}^{0} \right)=\dfrac{{{2}^{n+1}}-1}{n+1}$.

Chọn D

Ta có: $kC_{n}^{k}=k.\dfrac{n!}{k!(n-k)!}=\dfrac{n!}{(k-1)!\text{ }\!\![\!\!\text{ }(n-1)-(k-1)\text{ }\!\!]\!\!\text{ }!}$

$=n\dfrac{(n-1)!}{(k-1)!\text{ }\!\![\!\!\text{ }(n-1)-(k-1)\text{ }\!\!]\!\!\text{ }!}=nC_{n-1}^{k-1}$, $\forall k\ge 1$

$\Rightarrow {{S}_{2}}=\sum\limits_{k=1}^{n}{nC_{n-1}^{k-1}}=n\sum\limits_{k=0}^{n-1}{C_{n-1}^{k}}=n{{.2}^{n-1}}$.

Chọn A

Ta có $k(k-1)C_{n}^{k}=\dfrac{n!}{(k-2)!(n-k)!}=n(n-1)C_{n-2}^{k-2}$

$\Rightarrow {{S}_{3}}=n(n-1)\sum\limits_{k=2}^{n}{C_{n-2}^{k-2}}=n(n-1){{2}^{n-2}}$.

Chọn D

Ta có $S={{S}_{1}}-{{S}_{2}}$, trong đó
${{S}_{1}}=C_{n}^{0}+\dfrac{{{3}^{2}}}{2}C_{n}^{1}+\dfrac{{{3}^{3}}}{3}C_{n}^{2}+…+\dfrac{{{3}^{n+1}}}{n+1}C_{n}^{n}$

${{S}_{2}}=\dfrac{1}{2}C_{n}^{1}+\dfrac{1}{3}C_{n}^{2}+…+\dfrac{1}{n+1}C_{n}^{n}$

Ta có ${{S}_{2}}=\dfrac{{{2}^{n+1}}-1}{n+1}-1$

Tính ${{S}_{1}}=?$

Ta có: $\dfrac{{{3}^{k+1}}}{k+1}C_{n}^{k}={{3}^{k+1}}\dfrac{n!}{(k+1)!(n-k)!}$ $=\dfrac{{{3}^{k+1}}}{n+1}\dfrac{(n+1)!}{(k+1)!\text{ }\!\![\!\!\text{ }(n+1)-(k+1)\text{ }\!\!]\!\!\text{ !}}$$=\dfrac{{{3}^{k+1}}}{n+1}C_{n+1}^{k+1}$

$\Rightarrow {{S}_{1}}=\dfrac{1}{n+1}\sum\limits_{k=0}^{n}{{{3}^{k+1}}C_{n+2}^{k+1}}-2C_{n}^{0}$$=\dfrac{1}{n+1}\left( \sum\limits_{k=0}^{n+1}{{{3}^{k}}C_{n+1}^{k}}-C_{n}^{0} \right)-2C_{n}^{0}$$=\dfrac{{{4}^{n+1}}-1}{n+1}-2$.

Vậy $S=\dfrac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}-1$.

Chọn A

Ta có: $S={{S}_{1}}-{{S}_{2}}$

Trong đó ${{S}_{1}}=\sum\limits_{k=0}^{n}{C_{n}^{k}\dfrac{{{2}^{k+1}}}{k+1}};\text{ }{{S}_{2}}=\sum\limits_{k=0}^{n}{\dfrac{C_{n}^{k}}{k+1}}=\dfrac{{{2}^{n+1}}-1}{n+1}-1$

Mà $\dfrac{{{2}^{k+1}}}{k+1}C_{n}^{k}=\dfrac{{{2}^{k+1}}}{n+1}C_{n+1}^{k+1}$$\Rightarrow {{S}_{1}}=\dfrac{{{3}^{n+1}}-1}{n+1}-1$

Suy ra: $S=\dfrac{{{3}^{n+1}}-{{2}^{n+1}}}{n+1}$.

Chọn B

Đặt $S=\sum\limits_{k=1}^{2n+1}{{{(-1)}^{k-1}}.k{{.2}^{k-1}}C_{2n+1}^{k}}$

Ta có: ${{(-1)}^{k-1}}.k{{.2}^{k-1}}C_{2n+1}^{k}=={{(-1)}^{k-1}}.(2n+1){{.2}^{k-1}}C_{2n}^{k-1}$

Nên $S=(2n+1)(C_{2n}^{0}-2C_{2n}^{1}+{{2}^{2}}C_{2n}^{2}-…+{{2}^{2n}}C_{2n}^{2n})=2n+1$

Vậy $2n+1=2005\Leftrightarrow n=1002$.

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: $VT=n({{3}^{0}}{{.5}^{n-1}}C_{n-1}^{0}+{{3}^{1}}{{.5}^{n-2}}C_{n-1}^{1}+…+{{3}^{n-1}}{{5}^{0}}C_{n-1}^{n-1})$

$=n{{(5+3)}^{n-1}}=n{{.8}^{n-1}}$

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 $S=n(n-1)(C_{n-2}^{0}+C_{n-2}^{1}+C_{n-2}^{2}+…+C_{n-2}^{n-2})=n(n-1){{2}^{n-2}}$

Chọn A

Ta có:${{\left( x+1 \right)}^{n}}{{\left( 1+x \right)}^{n}}={{\left( x+1 \right)}^{2n}}$.

Vế trái của hệ thức trên chính là:

$\left( C_{n}^{0}{{x}^{n}}+C_{n}^{1}{{x}^{n-1}}+…+C_{n}^{n} \right)\left( C_{n}^{0}+C_{n}^{1}x+…+C_{n}^{n}{{x}^{n}} \right)$

Và ta thấy hệ số của ${{x}^{n}}$ trong vế trái là

${{\left( C_{n}^{0} \right)}^{2}}+{{\left( C_{n}^{1} \right)}^{2}}+{{\left( C_{n}^{2} \right)}^{2}}+…+{{\left( C_{n}^{n} \right)}^{2}}$

Còn hệ số của ${{x}^{n}}$ trong vế phải ${{\left( x+1 \right)}^{2n}}$ là $C_{2n}^{n}$

Do đó ${{\left( C_{n}^{0} \right)}^{2}}+{{\left( C_{n}^{1} \right)}^{2}}+{{\left( C_{n}^{2} \right)}^{2}}+…+{{\left( C_{n}^{n} \right)}^{2}}=C_{2n}^{n}$

Chọn D

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

Chọn D

Xét khai triển:

${{(1+x)}^{2011}}=C_{2011}^{0}+xC_{2011}^{1}+{{x}^{2}}C_{2011}^{2}+…+{{x}^{2010}}C_{2011}^{2010}+{{x}^{2011}}C_{2011}^{2011}$

Cho $x=2$ ta có được:

${{3}^{2011}}=C_{2011}^{0}+2.C_{2011}^{1}+{{2}^{2}}C_{2011}^{2}+…+{{2}^{2010}}C_{2011}^{2010}+{{2}^{2011}}C_{2011}^{2011}$ (1)

Cho $x=-2$ ta có được:

$-1=C_{2011}^{0}-2.C_{2011}^{1}+{{2}^{2}}C_{2011}^{2}-…+{{2}^{2010}}C_{2011}^{2010}-{{2}^{2011}}C_{2011}^{2011}$ (2)

Lấy (1) + (2) ta có:

$2\left( C_{2011}^{0}+{{2}^{2}}C_{2011}^{2}+…+{{2}^{2010}}C_{2011}^{2010} \right)={{3}^{2011}}-1$

Suy ra:${{S}_{2}}=C_{2011}^{0}+{{2}^{2}}C_{2011}^{2}+…+{{2}^{2010}}C_{2011}^{2010}=\dfrac{{{3}^{2011}}-1}{2}$.

Chọn B

Ta có: $kC_{n}^{k}=k.\dfrac{n!}{k!(n-k)!}=\dfrac{n!}{(k-1)!\text{ }\!\![\!\!\text{ }(n-1)-(k-1)\text{ }\!\!]\!\!\text{ }!}$
$=n\dfrac{(n-1)!}{(k-1)!\text{ }\!\![\!\!\text{ }(n-1)-(k-1)\text{ }\!\!]\!\!\text{ }!}=nC_{n-1}^{k-1}$, $\forall k\ge 1$

$\Rightarrow {{S}_{3}}=\sum\limits_{k=1}^{n}{nC_{n-1}^{k-1}}=n\sum\limits_{k=0}^{n-1}{C_{n-1}^{k}}=n{{.2}^{n-1}}$.