Bài 1: 

a: \(2A=2^{101}+2^{100}+...+2^2+2\)

\(\Leftrightarrow A=2^{100}-1\)

b: \(3B=3^{101}+3^{100}+...+3^2+3\)

\(\Leftrightarrow2B=3^{100}-1\)

hay \(B=\dfrac{3^{100}-1}{2}\)

c: \(4C=4^{101}+4^{100}+...+4^2+4\)

\(\Leftrightarrow3C=4^{101}-1\)

hay \(C=\dfrac{4^{101}-1}{3}\)

tính riêng:

\(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\)

=\(\left(\frac{100}{99}-1\right)+\left(\frac{100}{98}-1\right)+\left(\frac{100}{97}-1\right)+...+\left(\frac{100}{2}-1\right)+99\)

=\(100.\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)+99-98\) 

=\(100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)\)

vậy \(\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=100\)

chúc bạn học tốt ^^

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

3\(\times\) A  =  3  +  \(\dfrac{1}{3}\) +  \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+........+ \(\dfrac{1}{3^{n-1}}\)

3A - A =  3 + \(\dfrac{1}{3}\) - 1 - \(\dfrac{1}{3^n}\) 

    2A  = \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)

      A  = ( \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)): 2

     A =   \(\dfrac{7.3^{n-1}-1}{3^n}\) : 2

     A = \(\dfrac{7.3^{n-1}-1}{2.3^n}\)

   B   =      \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+......+\(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)

2B    =  2 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) -  \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)

2B + B = 2 - \(\dfrac{1}{2^{100}}\)

  3B     =  2 - \(\dfrac{1}{2^{100}}\)

    B     =   ( 2 - \(\dfrac{1}{2^{100}}\)): 3

    B     =     \(\dfrac{2.2^{100}-1}{2^{100}}\) : 3

    B     = \(\dfrac{2^{101}-1}{3.2^{100}}\)

-Quy luật: Nhân mỗi vế của đẳng thức cho số thích hợp để tạo ra đẳng thức mới, khi cộng (hoặc trừ) mỗi vế của mỗi đẳng thức thì sẽ rút gọn bớt.

a) \(A=2-2^2+2^3-2^4+...+2^{99}-2^{100}\)

\(\Rightarrow2A=2^2-2^3+2^4-2^5+...+2^{100}-2^{101}\)

\(\Rightarrow2A+A=2^2-2^3+2^4-2^5+...+2^{100}-2^{101}+\left(2-2^2+2^3-2^4+...+2^{99}-2^{100}\right)\)

\(\Rightarrow A=-2^{101}+2\)

b,c) làm tương tự. 

d) \(D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow3D=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow3D-D=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow2D=3+\dfrac{1}{3^{100}}\)

\(\Rightarrow2D=\dfrac{3^{101}+1}{3^{100}}\Rightarrow D=\dfrac{3^{101}+1}{2.3^{100}}\)

e) làm tương tự nhưng đổi thành cộng.

\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)

\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)

\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)

\(2C=1-\frac{1}{3^{99}}< 1\)

\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)

1.

B = 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1

3B = 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3

3B + B = ( 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3 ) + ( 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1 )

4B = 3¹⁰¹ + 1

B = \(\frac{3^{101}+1}{4}\)

\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)

\(=\frac{\left(101+1\right).100:2}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}\)

\(=\frac{5050}{1+1+...+1+1}\)(51 chữ số 1)

= \(\frac{5050}{51}\)