1.\(A=\left(-5\right)^0+\left(-5\right)^2+..+\left(-5\right)^{50}\)

\(\Rightarrow\left(-5\right)^2.A=\left(-5\right)^2+\left(-5\right)^4+..+\left(-5\right)^{52}=5^{52}-1+A\)

\(\Rightarrow A=\frac{5^{52}-1}{24}\)

b. Xét \(S=1+2+..+2^{2009}\Rightarrow2S=2+2^2+..+2^{2010}=2^{2010}+S-1\)

\(\Leftrightarrow S=2^{2010}-1\Rightarrow B=1\)

\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(=\frac{1}{1.2}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)

Bg

Đặt A = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)

=> A = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

=> A = \(\frac{1}{1.2}-\frac{1}{99.100}\)

=> A = \(\frac{1}{2}-\frac{1}{9900}\)

=> A = \(\frac{4949}{9900}\)

Vậy giá trị của biểu thức trên là \(\frac{4949}{9900}\)

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\)

=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)

Khi đó \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\right)\)

=> \(A=1-\frac{1}{2^{2013}}< 1\left(\text{Đpcm}\right)\)

k=1 nên x =3, y=4 , z=1