là sao

Ta có: \(\frac{1}{201\times299}+\frac{1}{203\times297}+\cdots+\frac{1}{297\times203}+\frac{1}{299\times201}\)

\(=\frac{2}{201\times299}+\frac{2}{203\times297}+\cdots+\frac{2}{249\times251}\)

\(=\frac{2}{500}\times\left(\frac{500}{201\times299}+\frac{500}{203\times297}+\cdots+\frac{500}{249\times251}\right)\)

\(=\frac{1}{250}\times\left(\frac{1}{201}+\frac{1}{299}+\frac{1}{203}+\frac{1}{297}+\cdots+\frac{1}{249}+\frac{1}{251}\right)\)

\(=\frac{1}{250}\times\left(\frac{1}{201}+\frac{1}{203}+\cdots+\frac{1}{297}+\frac{1}{299}\right)\)

Ta có: \(\frac{\left(\frac{1}{201}+\frac{1}{203}+\cdots+\frac{1}{297}+\frac{1}{299}\right)}{\frac{1}{201\times299}+\frac{1}{203\times297}+\cdots+\frac{1}{297\times203}+\frac{1}{299\times201}}\)

\(=\frac{\left(\frac{1}{201}+\frac{1}{203}+\cdots+\frac{1}{297}+\frac{1}{299}\right)}{\frac{1}{250}\times\left(\left(\frac{1}{201}+\frac{1}{203}+\cdots+\frac{1}{297}+\frac{1}{299}\right)\right)}\)

\(=1:\frac{1}{250}=250\)

( 299 - 301 ) + ( 298 - 300 ) + ( 297 - 299 ) + ( 1 - 3 )

=    -2           +        -2   +        -2          +    -2

=                  -8

ta co: (299-301)+(298-300)+(297-299)...+(1/3)

(tong tren co 299-1+1=299 so hang )

=(-2)+ (-2)+(-2)+(-2)+...+(-2) =(-2).299:2=(-299)

k nha

203

{–(-299) -299} +203

= 0 + 203

= 203

–(-299)-299}+203

=0+203

=203

\(A=2+2^3+...+2^{101}\)

\(4A=2^3+2^5+...+2^{101}+2^{103}\)

\(4A-A=2^{103}-2\)

\(3A=2^{103}-2\)

\(A=\dfrac{2^{103}-2}{3}\)

\(\Rightarrow1+2+2^3+...+2^{101}=A+1=\dfrac{2^{103}+1}{3}\)

`[` Tham Khảo `]`

`( 299/6 +1/2) xx 50 :2`

`= (299/6 + 3/6) xx 50 xx 1/2`

`= 302/6 xx 50 xx 1/2`

`=151/3 xx 50 xx 1/2`

`=7550/6`

`=3775/3`

Đặt :

\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{99}}\)

\(\Leftrightarrow2A=3+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{98}}\)

\(\Leftrightarrow2A-A=\left(3+\dfrac{1}{2}+....+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{2^{99}}\right)\)

\(\Leftrightarrow A=2-\dfrac{1}{2^{99}}\)

Vậy..