2 điểm!?

thi hay sao?

S= 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

S= 3 x ( 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 )

S = 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

S= 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 - 1 - 1/9 -1/27 - 1/81 - 1/243 - 1/729

S = 3 - 1/729 

S= 142/729

a. Dãy là tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=\dfrac{1}{10}\end{matrix}\right.\)

Do đó: \(S=\dfrac{u_1}{1-q}=\dfrac{1}{1-\dfrac{1}{10}}=\dfrac{10}{9}\)

b. Tương tự, tổng cấp số nhân lùi vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=-\dfrac{1}{3}\end{matrix}\right.\) bạn tự ráp công thức

c. \(S=2+S_1\) với \(S_1\) là cấp số nhân lùi vô hạn \(\left\{{}\begin{matrix}u_1=\dfrac{3}{10}\\q=\dfrac{3}{10}\end{matrix}\right.\)

\(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\)

\(=3\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\)

Đặt \(A=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2A=2+1+...+\dfrac{1}{2^8}\)

\(\Rightarrow2A-A=\left(2+1+...+\dfrac{1}{2^8}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\right)\)

\(\Rightarrow A=2-\dfrac{1}{2^9}\)

\(\Rightarrow S=3\left(2-\dfrac{1}{2^9}\right)=6-\dfrac{3}{2^9}\)

Vậy...

=> S = 3 \(\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\right)\)

S = 3 \(\left(2+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2^2}+...+\dfrac{1}{2^8}-\dfrac{1}{2^9}\right)\)

= 3\(\left(2-\dfrac{1}{2^9}\right)\)= 6 \(-\dfrac{3}{2^9}\)

TICK CHO MK NHA!

Ta có:

S = 3 + \(\dfrac{3}{2}\) + \(\dfrac{3}{2^2}\) + ........+ \(\dfrac{3}{2^9}\)

2S = 6 + 3 + \(\dfrac{3}{2^2}\) + ....... + \(\dfrac{3}{2^8}\)

Mà S = 3 + \(\dfrac{3}{2}\) + \(\dfrac{3}{2^2}\) + ........+ \(\dfrac{3}{2^9}\)

=> 2S - S = 6 - \(\dfrac{3}{2^9}\)

\(S=3\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\)

Đặt : \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\\ \Rightarrow2.A=2+1+....+\dfrac{1}{2^8}\\ \Rightarrow A=2-\dfrac{1}{2^9}\\ \Rightarrow S=3\left(2-\dfrac{1}{2^9}\right)\)

Ta có :

\(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+.........+\dfrac{3}{2^9}\)

\(2S=6+3+\dfrac{3}{2}+\dfrac{3}{2^2}+..............+\dfrac{3}{2^8}\)

\(2S-S=\left(6+3+\dfrac{3}{2}+..........+\dfrac{3}{2^8}\right)-\left(3+\dfrac{3}{2}+\dfrac{3}{2^2}+............+\dfrac{3}{2^9}\right)\)

\(S=6-\dfrac{3}{2^9}\)

\(S=\dfrac{6.512}{512}-\dfrac{3}{512}\)

\(S=\dfrac{3069}{512}\)

S   = 1/3 + 1/3^2 + 1/3^3 + 1/3^4 + ... + 1/3^99 + 1/3^100

3S = 1 +1/3 +1/3^2 +1/3^3 + ... + 1/3^98 +1/3^99

3S - S = ( 1 + 1/3 + 1/3^2 +1/^3 + ... + 1/3^98 +1/3^99 ) - ( 1/3 + 1/3^2 + 1/3^3 + 1/3^4 +... + 1/3^99 + 1/3^100 )

2S = 1 - 1/3^100

S   = (1 - 1/3^100). 1/2

Ta có: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)

\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

Ta có S = \(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\dfrac{4}{16}+...+\dfrac{10}{2^{10}}\)

             = \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{10}{2^{10}}\)

2S = 1 + \(\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{10}{2^9}\)

2S - S = ( 1 + \(\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{10}{2^9}\)) - ( \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{10}{2^{10}}\))

S = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}-\dfrac{10}{2^{10}}\)

Đặt A = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)

2A = 2 + 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)

2A - A = ( 2 + 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)) - ( 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\))

A = 2 - \(\dfrac{1}{2^9}\)

⇒ S = 2 - \(\dfrac{1}{2^9}\) - \(\dfrac{10}{2^{10}}\) = \(\dfrac{2^{11}}{2^{10}}-\dfrac{2}{2^{10}}-\dfrac{10}{2^{10}}=\dfrac{2^2\left(2^9-3\right)}{2^{10}}=\dfrac{2^9-3}{2^8}\)

Vậy S = \(\dfrac{2^9-3}{2^8}\)

Tổng quát:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow S=\dfrac{10}{11}\)

Ta có công thức tổng quát như sau:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\right]\left[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}\right]}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)

\(=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)

\(=\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}\)

Áp dụng vào tổng S ta có:

\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{121\sqrt{120}+120\sqrt{121}}\)

\(S=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{120}}+\dfrac{1}{\sqrt{121}}\)

\(S=1-\dfrac{1}{\sqrt{121}}=1-\dfrac{1}{11}=\dfrac{10}{11}\)