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 \(C=\dfrac{4^{101}-1}{3}\)

A= \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)

2A= \(2.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\right)\)

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

⇒ 2A- A= \(1-\dfrac{1}{2^{100}}\)

⇒ A= \(1-\dfrac{1}{2^{100}}\)

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

3B= \(3.\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\)

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

⇒ 3B- B= \(1-\dfrac{1}{3^{100}}\)

⇒ B.(3-1)= \(1-\dfrac{1}{3^{100}}\)

⇒ 2B= \(1-\dfrac{1}{3^{99}}\)

⇒ B= \(\left(1-\dfrac{1}{3^{99}}\right):2\)

⇒ B= \(\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)

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

=>\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

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

1/2.A=1/2²+1/2³+...+1/2¹⁰¹

=>1/2A-A=1/2¹⁰¹-1/2

=>-1/2A=1/2¹⁰¹-1/2

A=(1/2¹⁰¹-1/2):(-1/2)=(1/2¹⁰¹-1/2).(-2)

=1-1/2¹⁰⁰

`A=sqrt{1+1/a^2+1/(a+1)^2}`
`=sqrt{1/a^2+2/a+1-2/a+1/(a+1)^2}`
`=sqrt{(1/a+1)^2-2/a+1/(a+1)^2}`
`=sqrt{(a+1)^2/a^2-2.(a+1)/a.(1/(a+1))+1/(a+1)^2}`
`=sqrt{((a+1)/a-1/(a+1))^2}`
`=|(a+1)/a-1/(a+1)|`
`=|1+1/a-1/(a+1)|`
`a>0=>1/a>1/(a+1)=>1+1/a-1/(a+1)>0`
`=>A=1+1/a-1/(a+1)`

Áp dụng công thức ở A ta tính được

`B=1+1/1-1/2+1+1/2-1/3+1-1/3+1/4+.......+1+1/(n-1)-1/n`(ở sau bạn không ghi rõ nên mình đặt số cuối là n)

`=underbrace{1+1+....+1}_{\text{n chữ số 1}}-1/n`

`=n-1/n`