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

Đặt \(B=2^{99}+2^{98}+...+2+1\)

\(\Rightarrow2B=2^{100}+2^{99}+...+2^2+2\)

\(\Rightarrow2B-B=2^{100}-1\Leftrightarrow B=2^{100}-1\)

\(\Rightarrow A=2^{100}-\left(2^{100}-1\right)=1\)

a) \(A=1+2+2^2+...+2^{50}\)

\(\Rightarrow2A=2+2^2+...+2^{51}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)

b) \(B=1+3+3^2+...+3^{100}\)

\(\Rightarrow3B=3+3^2+...+3^{101}\)

\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)

\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)

c) \(C=5+5^2+...+5^{30}\)

\(\Rightarrow5C=5^2+5^3+...+5^{31}\)

\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)

\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)

d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)

\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)

\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)

\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)

??????

1+2-3-4+5+6-7-8+9+10-11-12+........+298-299-300+301+302 = 

1+2+(5-3)+(6-4)+(9-7)+(10-8)+…….+(301-299)+(302-300)=

Từ 302 đến 3 có số cặp là [(302-3):1+1]:2=150 cặp. Mà mỗi cặp có giá trị là 2

Vậy 1+2-3-4+5+6-7-8+9+10-11-12+........+298-299-300+301+302 = 

1+2+2×150=3+300=303

và 1 + 1 = ?:D

*Ý 1 :Áp dụng công thức tính nhanh dãy phân số, ta làm như sau:

Lấy 2.B như sau:

\(2.B=2.\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)

\(\Leftrightarrow2.B=2.\frac{1}{2}+2.\frac{1}{2^2}+...+2.\frac{1}{2^{98}}+2.\frac{1}{2^{99}}\)

\(\Leftrightarrow2.B=1+\frac{1}{2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)

Ta thấy: \(2.B\)và \(B\)cùng có \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)

Nên lấy \(2.B-B\)ta sẽ có:

\(\Rightarrow2.B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)

\(\Leftrightarrow B=1-\frac{1}{2^{99}}\)

Vậy tổng \(B=1-\frac{1}{2^{99}}.\)

* Ý 2:\(\left|3x+5\right|=10\)

\(\Rightarrow\orbr{\begin{cases}3x+5=10\\3x+5=-10\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-5\end{cases}}}\)

Vậy \(x\in\left\{-5;\frac{5}{3}\right\}.\)

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..

Ta có: \(A=2^{100}-2^{99}-2^{98}-...-2^2-2-1\)

\(\Leftrightarrow2A=2^{101}-2^{100}-2^{99}-...-2^3-2^2-2\)

\(\Leftrightarrow2A-A=2^{101}-2^{100}-2^{99}-...-2^3-2^2-2-2^{100}+2^{99}+2^{98}+...+2^2+2+1\)

\(\Leftrightarrow A=2^{101}-2\cdot2^{100}+1\)

\(\Leftrightarrow A=1\)

Thank you