Đặt A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{2013}{2014}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)

\(\Rightarrow A=1-\frac{1}{x+1}=\frac{2013}{2014}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{2013}{2014}\)

\(\Rightarrow\)\(\frac{1}{x+1}=\frac{1}{2014}\)

\(\Rightarrow x+1=2014\)

\(\Rightarrow x=2014-1\)

\(\Rightarrow x=2013\)

Vậy x=2013

 \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)

\(1-\frac{1}{x+1}=\frac{2013}{2014}\)

\(\frac{1}{x+1}=1-\frac{2013}{2014}\)

\(\frac{1}{x+1}=\frac{1}{2014}\)

Vì \(x+1\)là mẫu số nên:

\(x+1=2014\)

\(x=2014-1=2013\)

Vậy ....

  P/s: Dấu . là nhân nha!

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{n\left(n+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(=1-\frac{1}{n+1}\)

\(=\frac{n+1}{n+1}-\frac{1}{n+1}\)

\(=\frac{n}{n+1}\)

cac cau tu di ma lam bai tap cua minh

cau len mang di , bai nay mk chua hoc , sory nha

chuc ban hoc tot ^-^

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=1-\dfrac{1}{25}\)

\(=\dfrac{24}{25}\)

Nhanh giúp mình với cả nhà ơi

Đặt A = 1x2+2x3+3x4+...+nx(n+1)

=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... + n.(n + 1).[(n + 2).(n - 1)]

=> 3A =  1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + n.(n + 1).(n + 2)

=> 3A = n.(n + 1).(n + 2)

=> A = n.(n + 1).(n + 2) / 3

Cách làm mk làm giống  Edokawa Conan nhé kw ;\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)

program tinhtoan;

uses crt;

var: i;n:interger;

S:real;

writeln(' Nhap n='); readln(n);

S:=0;

For i:=1 to n*(n*1) do S:=S+\(\frac{1}{i};\)

writeln(' S=',S);

End.

(ps: ko chắc )

\(C=1.2+2.3+3.4+...+n\left(n+1\right)\\ \Rightarrow3.C=1.2.3+2.3.3+3.4.3+..+n\left(n+1\right).3\\ \Rightarrow3.C=1.2.3+2.3.4-1.2.3+....+n\left(n+1\right)\left(n+2\right)-\left(n-1.n.\left(n+1\right)\right)\\ \Rightarrow3.C=n\left(n+1\right)\left(n+2\right)\\ \Rightarrow C=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)

Cái D tính TT

a)

Số số hạng của dãy trên là;

     (n - 1) : 1 + 1 = n(số hạng)

Tổng dãy trên là:

       (n + 1) x n : 2 = ? (tùy giá trị n)

b) Đặt A =  1x2 + 2x3 + 3x4 + ... + 99 x 100

3A= 3 x ( 1x2 + 2x3 + 3x4 + ... + 99 x 100)

3A = 1 x 2 x (3 - 0) + 2 x 3 x(4-1) + .....+99.100.(101 - 98)

3A = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 + .......+ 99.100.101

3A = 99.100.101

A   = \(\frac{\text{99.100.101}}{3}=333300\)

a, 1 + 2 + 3 + ... + n

= ( 1 + n) × n : 2

b, 1×2 + 2×3 + 3×4 + ... + 99×100

= 1/3 × ( 1×2×3 + 2×3×3 + 3×4×3 + ... + 99×100×3)

= 1/3 × [ 1×2×(3-0) + 2×3×(4-1) + 3×4×(5-2) + ... + 99×100×(101-98) ]

= 1/3 × ( 1×2×3 - 0×1×2 + 2×3×4 - 1×2×3 + 3×4×5 - 2×3×4 + ... + 99×100×101 - 98×99×100 )

= 1/3 × [ ( 1×2×3 + 2×3×4 + 3×4×5 + ... + 99×100×101) - ( 0×1×2 + 1×2×3 + 2×3×4 + ... + 98×99×100) ]

= 1/3 × ( 99×100×101 - 0×1×2)

= 1/3 × ( 99×100×101 - 0)

= 1/3 × 99×100×101

= 333 300