a) \(A=2+2^2+...+2^{120}\)

\(\Rightarrow2A=2^2+2^3+...+2^{121}\)

\(\Leftrightarrow2A-A=\left(2^2+2^3+...+2^{121}\right)-\left(2+2^2+...+2^{120}\right)\)

\(\Rightarrow A=2^{121}-2\)

b) Mk làm mẫu 1 phần thôi nhé bn:

\(A=2+2^2+...+2^{120}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{119}\left(1+2\right)\)

\(A=3\left(2+2^3+...+2^{119}\right)\) chia hết cho 3

Tương tự xét chia hết cho 7 thì nhóm 3 số, cho 15 thì 4 số nhé

Ta có: \(C=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{100}}\)

\(\Rightarrow3C=1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{99}}\)

\(\Rightarrow3C+C=\left(1-\frac{1}{3}+...-\frac{1}{3^{99}}\right)+\left(\frac{1}{3}-\frac{1}{3^2}+...-\frac{1}{3^{100}}\right)\)

\(\Leftrightarrow4C=1-\frac{1}{3^{100}}\)

\(\Rightarrow C=\frac{3^{100}-1}{4\cdot3^{100}}\)

Ta có: \(B=\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+...+\frac{1}{3^{99}}\)

\(\Rightarrow9B=3+\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{97}}\)

\(\Rightarrow9B-B=\left(3+\frac{1}{3}+...+\frac{1}{3^{97}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(\Leftrightarrow8B=3-\frac{1}{3^{99}}\)

\(\Rightarrow B=\frac{3^{100}-1}{8\cdot3^{99}}\)

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

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)

\(\Leftrightarrow2A=1-\frac{1}{3^{100}}\)

\(\Rightarrow A=\frac{3^{100}-1}{2\cdot3^{100}}\)

Ta có: \(1+2^2+3^2+4^2+...+99^2+100^2\) (đề đúng)

\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)+100\left(101-1\right)\)

\(=\left(1.2+2.3+3.4+...+99.100+100.101\right)-\left(1+2+3+...+100\right)\)

\(=\frac{1.2.3+2.3.3+...+100.101.3}{3}-\frac{\left(100+1\right)\left[\left(100-1\right)\div1+1\right]}{2}\)

\(=\frac{1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+100.101.\left(102-99\right)}{3}-5050\)

\(=\frac{1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-99.100.101+100.101.102}{3}-5050\)

\(=\frac{100.101.102}{3}-5050\)

\(=343400-5050\)

\(=338350\)

\(\sqrt{x}+1=40\Rightarrow\sqrt{x}=39\Rightarrow\left(\sqrt{x}\right)^2=39^2\Rightarrow x=1521\)

\(3\sqrt{x}+1=40\)

ĐK : x ≥ 0

<=> \(3\sqrt{x}=39\)

<=> \(\sqrt{x}=13\)

<=> \(x=169\)( tm )

Vậy x = 169

\(a,\frac{34}{5}:\frac{8}{5}=0,25:x\)

\(\frac{17}{4}=\frac{1}{4}:x\)

\(x=\frac{1}{4}:\frac{17}{4}\)

\(x=\frac{1}{17}\)

\(b,2x+\frac{3}{24}=3x-\frac{1}{32}\)

\(2x-3x=\frac{-1}{32}-\frac{1}{8}\)

\(-x=\frac{-5}{32}\)

\(x=\frac{5}{32}\)

\(13x-\frac{2}{2}x+5=\frac{76}{17}\)

\(12x=\frac{-9}{17}\)

\(x=\frac{-3}{68}\)

a,\(\frac{34}{5}\div\frac{8}{5}=0,25\div x\)

\(\frac{17}{4}=0,25\div x\)

\(x=17\)

b,\(2x+\frac{3}{24}=3x-\frac{1}{32}\)

\(\Leftrightarrow2x-3x=-\frac{1}{32}-\frac{3}{24}\)

\(\Leftrightarrow2x-3x=-\frac{5}{32}\)

\(\Leftrightarrow-x=-\frac{5}{32}\)

\(\Leftrightarrow x=\frac{5}{32}\)