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a: \(=625\cdot3-4\cdot\left(15^4-1\right)\)

\(=1875-4\cdot50625+4\)

\(=1879-202500=-200621\)

b: =50+49+48+47+...+2+1

Số số hạng là 50-1+1=50(số)

Tổng của dãy số là:

\(\dfrac{\left(1+50\right)\cdot50}{2}=51\cdot25=1275\)

28 tháng 7 2016

a. 5^4.3-4.(15^4-1)

=5^4.3-4.15^4+4

=4-15^4

=-50621

a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)

\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)

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

=>B/A=1/100

b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)

\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)

\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)

=>A/B=25

9 tháng 4 2017

Bài 1:

Ta có:

\(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\)

\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}\)

\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+...+\left(1+\dfrac{48}{2}\right)+1}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)

Vậy \(\dfrac{S}{P}=\dfrac{1}{50}\)

Bài 2:

Ta có:

\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)

\(=\dfrac{1}{5}+\left(\dfrac{1}{9}+\dfrac{1}{10}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}\right)\)

Nhận xét:

\(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\)

\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

Vậy \(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{2}\)