E = -1/3 +1/(3^2) - 1/(3^3) + .... - 1/(3^51)

E.1/3 = -1/(3^2) + 1/(3^3)-1/(3^4) +.... - 1/(3^52)

E + E.1/3 = [-1/3+1/(3^2) - 1/(3^3) +.... -1/(3^51)]+[-1/(3^2) +1/(3^3) -1/(3^4) +.... - 1/(3^52)]

E.4/3 = -1/3-1/(3^52)

E.4/3 = (-3^51 - 1)/(3^52)

E = (-3^51 - 1)/(3^52) . 3/4

E = (-3^51-1)/(4.3^51)

Ta có : 

\(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)

= \(\left(1\frac{1}{2}+\frac{1}{2}\right)+\left(2\frac{2}{3}+\frac{1}{3}\right)+\left(3\frac{3}{4}+\frac{1}{4}\right)+...+\left(49\frac{49}{50}+\frac{1}{50}\right)+\left(50\frac{50}{51}+\frac{1}{51}\right)\)

= \(2+3+4+5+...+49+50+51\)

= \(\left(\frac{51-2}{1}+1\right).\frac{51+2}{2}\)

= \(50.26,5\)

= 1325

Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+...+\dfrac{1}{3^{47}}+\dfrac{1}{3^{49}}+\dfrac{1}{3^{51}}\)

Và \(B=\dfrac{1}{3^2}+\dfrac{1}{3^4}+\dfrac{1}{3^6}+...+\dfrac{1}{3^{46}}+\dfrac{1}{3^{48}}+\dfrac{1}{3^{50}}\)

Ta có:

\(9A=3+\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{45}}+\dfrac{1}{3^{47}}+\dfrac{1}{3^{49}}\)

\(9A-A=\left(3+\dfrac{1}{3}+...+\dfrac{1}{3^{47}}+\dfrac{1}{3^{49}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{49}}+\dfrac{1}{3^{51}}\right)\)

\(8A=3-\dfrac{1}{3^{51}}\)

\(A=\dfrac{3-\dfrac{1}{3^{51}}}{8}\)

\(9B=1+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{44}}+\dfrac{1}{3^{46}}+\dfrac{1}{3^{48}}\)

\(9B-B=\left(1+\dfrac{1}{3^2}+...+\dfrac{1}{3^{46}}+\dfrac{1}{3^{48}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{48}}+\dfrac{1}{3^{50}}\right)\)

\(8B=1-\dfrac{1}{3^{50}}\)

\(B=\dfrac{1-\dfrac{1}{3^{50}}}{8}\)

Suy ra

\(-\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{50}}-\dfrac{1}{3^{51}}=B-A=\dfrac{1-\dfrac{1}{3^{50}}}{8}-\dfrac{3-\dfrac{1}{3^{51}}}{8}\)

\(=\dfrac{\left(1-\dfrac{1}{3^{50}}\right)-\left(3-\dfrac{1}{3^{51}}\right)}{8}=\dfrac{-2-\dfrac{1}{3^{50}}+\dfrac{1}{3^{51}}}{8}=\dfrac{-2+\dfrac{-3^{51}+3^{50}}{3^{101}}}{8}\)

\(=\dfrac{-2+\dfrac{3^{50}\left(-3+1\right)}{3^{101}}}{8}=\dfrac{-2-\dfrac{2}{3^{51}}}{8}=-\dfrac{2\left(1+\dfrac{1}{3^{51}}\right)}{8}=-\dfrac{1+\dfrac{1}{3^{51}}}{4}\)

E=-1/3+1/3^2-1/3^3+1/3^4-...+1/3^50-1/3^51

3E=-1+1^2-1^3+1^4-1^5+...+1^50-1^51

3E=-1+1-1+1-1+...+1-1

3E=0

mình thiếu

bổ sung:

E=0:3

E=0

đặt tổng là S => S= \(\frac{-1}{3}+\left(\frac{-1}{3}\right)^2+\left(\frac{-1}{3}\right)^3+...+\left(\frac{-1}{3}\right)^{51}\)

xét tổng S'= a+ a²+... +a⁵¹ 

a.S'= a²+ ..+ a⁵² = S'-a+ a⁵² => S' = \(\frac{a^{52}-a}{a-1}\).

Thay  a= \(\frac{-1}{3}=>S=\frac{\left(\frac{-1}{3}\right)^{52}+\frac{1}{3}}{\frac{-1}{3}-1}=-\frac{\frac{1}{3^{51}}+1}{4}=-\frac{1}{4.3^{51}}-\frac{1}{4}\)

mình là công trừ cộng trừ mà bạn

\(+\frac{-1}{3}\) và \(-\frac{1}{3}\)khác nhau không bạn

-----A=-1/3+1/3^2-1/3^3+-----+1/3^50-1/... 
A*1/3 =-1/3^2+1/3^3+--------------------+1/3^5...  
--------A=-1/3+1/3^2-1/3^3+...+1/3^50-1... 
-------A*1/3 =-1/3^2+1/3^3+..---------...+1/3^51-1/3^... 
---------------------------------------... 
A+A*1/3=-1/3+0...+0+...0---------------... 
A+A*1/3= -1/3-1/3^52 
4/3*A= -1/3-1/3^52 
Vậy 
A= -(1/3+1/3^52)*3/4. 

Từ dãy trên ta có:

(\(\frac{3}{2}\)+\(\frac{1}{2}\))+(\(\frac{8}{3}\)+\(\frac{2}{3}\))+......+(\(\frac{2600}{51}\)+\(\frac{1}{51}\))                  < vì không có cách nhập hỗn số nên mình đổi ra phân số >

= 2 + 3 + 4 + 5 + 6 + ..........................+ 51

Từ 2 -> 51 có :( 51 - 2 ) : 1 + 1 = 50 số 

Chia ra : 50 : 2 = 25 cặp 

ta có( 51 + 2 ) x 25 =1325

Vậy tổng trên có kết quả bằng 1325       (tớ chỉ nghĩ thế thôi chứ sai đừng trách nhá.Đùa thôi,đúng đấy )