a)\(2S=2\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)\)

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

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

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

phần b tương tự

a. S=1+1/2+1/2^2+1/2^3+...+1/2^100

2S=2+1+1/2+1/2^2+...+1/2^99

2S-S=(2+1+1/2+1/2^2+...+1/2^99)-(1+1/2+1/2^2+1/2^3+...+1/2^100)

S=2-1/2^100

S=2^101-1/2^100

\(S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{299}+3^{300}\right)\\ S=\left(1+3\right)\left(1+3^2+...+3^{299}\right)\\ S=4\left(1+3^2+...+3^{299}\right)⋮4\)

mơn mà như vậy là chx đủ đâu

Ăn đầu buồi

giup minh voi sap phai nop roi

câu a Achia hết cho 128

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

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

     \(=1-\frac{1}{100}=\frac{99}{100}\)

Ta có: \(S=\frac{1}{5.6}+\frac{2}{6.8}+\frac{3}{8.11}+\frac{4}{11.15}+\frac{5}{15.20}\)

\(\Rightarrow S=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{15}-\frac{1}{20}\)

\(=\frac{1}{5}-\frac{1}{20}=\frac{4}{20}-\frac{1}{20}=\frac{3}{20}\)

2 câu c,d làm tương tự  

5:

a: \(3^{2n}=\left(3^2\right)^n=9^n\)

\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)

=>\(3^{2n}>2^{3n}\)

b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)

\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)

mà \(1568239201< 8036054027\)

nên \(199^{20}< 2003^{15}\)

4: \(100< 5^{2x-1}< 5^6\)

mà \(25< 100< 125\)

nên \(125< 5^{2x-1}< 5^6\)

=>3<2x-1<6

=>4<2x<7

=>2<x<7/2

mà x nguyên

nên x=3