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

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

Câu 1: Tổng $T=~\,\,C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}+…+C_{n}^{n}$ bằng:

[A]. $T\text{ }=\text{ }{{2}^{n}}$.

[B]. $T\text{ }=\text{ }{{2}^{n}}\text{ }1$.

[C]. $T\text{ }=\text{ }{{2}^{n}}+\text{ }1$.

[D]. $T\text{ }=\text{ }{{4}^{n}}$.

Hướng dẫn

Chọn A

Tính chất của khai triển nhị thức Niu – Tơn.

[Ẩn HD]

Câu 2: Tính giá trị của tổng $S\text{ }=\,\,C_{6}^{0}+C_{6}^{1}+..+C_{6}^{6}$ bằng:

[A]. 64.

[B]. 48.

[C]. 72.

[D]. 100.

Hướng dẫn

Chọn A

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

[Ẩn HD]

Câu 3: Khai triển ${{\left( x+y \right)}^{5}}$rồi thay $x,y$ bởi các giá trị thích hợp. Tính tổng $S=\,\,C_{5}^{0}+C_{5}^{1}+…+C_{5}^{5}$

[A]. 32.

[B]. 64.

[C]. 1.

[D]. 12.

Hướng dẫn

Chọn A

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

[Ẩn HD]

Câu 4: Tìm số nguyên dương n sao cho: $C_{n}^{0}+2C_{n}^{1}+4C_{n}^{2}+…+{{2}^{n}}C_{n}^{n}=243$

[A]. 4

[B]. 11

[C]. 12

[D]. 5

Hướng dẫn

Chọn D

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

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 ${{3}^{n}}=243={{3}^{5}}\Rightarrow n=5$.

[Ẩn HD]

Câu 5: Khai triển ${{\left( x+y \right)}^{5}}$rồi thay $x,y$ bởi các giá trị thích hợp. Tính tổng $S=\,\,C_{5}^{0}+C_{5}^{1}+…+C_{5}^{5}$

[A]. 32.

[B]. 64.

[C]. 1.

[D]. 12.

Hướng dẫn

Chọn A

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

[Ẩn HD]

Câu 6: 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 ${{a}_{10}}$.

[A]. ${{a}_{10}}=C_{5}^{0}.+C_{5}^{4}+C_{5}^{4}C_{5}^{3}$

[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 $S={{a}_{0}}-{{a}_{1}}+{{a}_{2}}-…-{{a}_{15}}$

[A]. 131

[B]. 147614

[C]. 0

[D]. 1

Hướng dẫn

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

[Ẩn HD]

Câu 7: 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 ${{a}_{4}}$

[A]. ${{a}_{4}}=C_{10}^{0}{{.2}^{4}}$

[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 $S={{a}_{1}}+2{{a}_{2}}+4{{a}_{3}}+…+{{2}^{20}}{{a}_{20}}$

[A]. $S={{17}^{10}}$

[B]. $S={{15}^{10}}$

[C]. $S={{17}^{20}}$

[D]. $S={{7}^{10}}$

Hướng dẫn

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 $S=f(2)={{17}^{10}}$

[Ẩn HD]

Câu 8: Tính tổng sau: $S=\dfrac{1}{2}C_{n}^{0}-\dfrac{1}{4}C_{n}^{1}+\dfrac{1}{6}C_{n}^{3}-\dfrac{1}{8}C_{n}^{4}+…+\dfrac{{{(-1)}^{n}}}{2(n+1)}C_{n}^{n}$

[A]. $\dfrac{1}{2(n+1)}$

[B]. 1

[C]. 2

[D]. $\dfrac{1}{(n+1)}$

Hướng dẫn

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

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

[Ẩn HD]

Câu 9: Tính tổng sau: $S=C_{n}^{1}{{3}^{n-1}}+2C_{n}^{2}{{3}^{n-2}}+3C_{n}^{3}{{3}^{n-3}}+…+nC_{n}^{n}$

[A]. $n{{.4}^{n-1}}$

[B]. 0

[C]. 1

[D]. ${{4}^{n-1}}$

Hướng dẫn

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}}$.

[Ẩn HD]

Câu 10: Tính các tổng sau: ${{S}_{1}}=C_{n}^{0}+\dfrac{1}{2}C_{n}^{1}+\dfrac{1}{3}C_{n}^{2}+…+\dfrac{1}{n+1}C_{n}^{n}$

[A]. $\dfrac{{{2}^{n+1}}+1}{n+1}$

[B]. $\dfrac{{{2}^{n+1}}-1}{n+1}$

[C]. $\dfrac{{{2}^{n+1}}-1}{n+1}+1$

[D]. $\dfrac{{{2}^{n+1}}-1}{n+1}-1$

Hướng dẫn

Chọn B

Ta có:

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

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

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

[Ẩn HD]

Câu 11: Tính các tổng sau:${{S}_{2}}=C_{n}^{1}+2C_{n}^{2}+…+nC_{n}^{n}$

[A]. $2n{{.2}^{n-1}}$

[B]. $n{{.2}^{n+1}}$

[C]. $2n{{.2}^{n+1}}$

[D]. $n{{.2}^{n-1}}$

Hướng dẫn

Chọn D

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

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

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

[Ẩn HD]

Câu 12: Tính các tổng sau:${{S}_{3}}=2.1.C_{n}^{2}+3.2C_{n}^{3}+4.3C_{n}^{4}+…+n(n-1)C_{n}^{n}$.

[A]. $n(n-1){{2}^{n-2}}$

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

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

[Ẩn HD]

Câu 13: Tính tổng $S=C_{n}^{0}+\dfrac{{{3}^{2}}-1}{2}C_{n}^{1}+…+\dfrac{{{3}^{n+1}}-1}{n+1}C_{n}^{n}$

[A]. $S=\dfrac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}$

[B]. $S=\dfrac{{{4}^{n+1}}+{{2}^{n+1}}}{n+1}-1$

 

[C]. $S=\dfrac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}+1$

[D]. $S=\dfrac{{{4}^{n+1}}-{{2}^{n+1}}}{n+1}-1$

Hướng dẫn

Chọn D

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

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

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

[Ẩn HD]

Câu 14: Tính tổng $S=C_{n}^{0}+\dfrac{{{2}^{2}}-1}{2}C_{n}^{1}+…+\dfrac{{{2}^{n+1}}-1}{n+1}C_{n}^{n}$

[A]. $S=\dfrac{{{3}^{n+1}}-{{2}^{n+1}}}{n+1}$

[B]. $S=\dfrac{{{3}^{n}}-{{2}^{n+1}}}{n+1}$

[C]. $S=\dfrac{{{3}^{n+1}}-{{2}^{n}}}{n+1}$

[D]. $S=\dfrac{{{3}^{n+1}}+{{2}^{n+1}}}{n+1}$

Hướng dẫn

Chọn A

Ta có: $S={{S}_{1}}-{{S}_{2}}$

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

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

[Ẩn HD]

Câu 15: Tìm số nguyên dương n sao cho : $C_{2n+1}^{1}-2.2C_{2n+1}^{2}+{{3.2}^{2}}C_{2n+1}^{3}-…+(2n+1){{2}^{n}}C_{2n+1}^{2n+1}=2005$

[A]. 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 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 $2n+1=2005\Leftrightarrow n=1002$.

[Ẩn HD]

Câu 16: Tính tổng${{1.3}^{0}}{{.5}^{n-1}}C_{n}^{n-1}+{{2.3}^{1}}{{.5}^{n-2}}C_{n}^{n-2}+…+n{{.3}^{n-1}}{{5}^{0}}C_{n}^{0}$

[A]. $n{{.8}^{n-1}}$

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

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

[Ẩn HD]

Câu 17: Tính tổng $S=2.1C_{n}^{2}+3.2C_{n}^{3}+4.3C_{n}^{4}+…+n(n-1)C_{n}^{n}$

[A]. $n(n+1){{2}^{n-2}}$

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

[Ẩn HD]

Câu 18: Tính tổng ${{\left( C_{n}^{0} \right)}^{2}}+{{\left( C_{n}^{1} \right)}^{2}}+{{\left( C_{n}^{2} \right)}^{2}}+…+{{\left( C_{n}^{n} \right)}^{2}}$

[A]. $C_{2n}^{n}$

[B]. $C_{2n}^{n-1}$

[C]. $2C_{2n}^{n}$

[D]. $C_{2n-1}^{n-1}$

Hướng dẫn

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

[Ẩn HD]

Câu 19: Tính tổng sau: ${{S}_{1}}={{5}^{n}}C_{n}^{0}+{{5}^{n-1}}.3.C_{n}^{n-1}+{{3}^{2}}{{.5}^{n-2}}C_{n}^{n-2}+…+{{3}^{n}}C_{n}^{0}$

[A]. ${{28}^{n}}$

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

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

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

Hướng dẫn

Chọn D

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

[Ẩn HD]

Câu 20: ${{S}_{2}}=C_{2011}^{0}+{{2}^{2}}C_{2011}^{2}+…+{{2}^{2010}}C_{2011}^{2010}$

[A]. $\dfrac{{{3}^{2011}}+1}{2}$

[B]. $\dfrac{{{3}^{211}}-1}{2}$

[C]. $\dfrac{{{3}^{2011}}+12}{2}$

[D]. $\dfrac{{{3}^{2011}}-1}{2}$

Hướng dẫn

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

[Ẩn HD]

Câu 21: Tính tổng ${{S}_{3}}=C_{n}^{1}+2C_{n}^{2}+…+nC_{n}^{n}$

[A]. $4n{{.2}^{n-1}}$

[B]. $n{{.2}^{n-1}}$

[C]. $3n{{.2}^{n-1}}$

[D]. $2n{{.2}^{n-1}}$

Hướng dẫn

Chọn B

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

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

[Ẩn HD]