Question

Cho A = \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+.....+\frac{2499}{2500}\)

CMR A không là số nguyên 

Answer 1

\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)

   \(=\frac{1.3}{2^2}+\frac{2.4}{3^2}+\frac{3.5}{4^2}+...+\frac{49.51}{50^2}\)

   \(=\frac{1.3.2.4.3.5...49.51}{2^2.3^2.4^2...50^2}\)

    \(=\frac{\left(1.2.3...49\right)\left(3.4.5...51\right)}{2^2.3^2.4^2...50^2}\)

    \(=\frac{1.2.50.51}{2^2.50^2}=\frac{51}{100}\)

  

Answer 2

đoạn thứ 3 bạn làm sao chuyển về như thế được Vimo Asdred?

Answer 3

51/100

Answer 4

lồn vãi

Answer 5

= \(\frac{51}{100}\)

Hk tốt

Answer 6

51/100 nhé vì mình lười viết

Answer 7

\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)

\(=\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{4^2}\right)+...+\left(1-\frac{1}{50^2}\right)\)

\(=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)< 49\)

Xét \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}< 1\)

\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>49-1=48\)

Lúc đó 48 < A < 49. Vậy A không là số nguyên