Question

[nguyen thi van khanh]

Tính \(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+4+...+50}\)

ai nhanh mk k cho 2 cái luôn

Answer 1

[Yến Nhi Libra Virgo HotGirl Sakura]

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+..+\frac{1}{1+2+3+...+50}\)

Ta có :

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

\(A=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{2550}\)

\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)

\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{51}\right)\)

\(A=2\times\frac{49}{102}\)

\(A=\frac{49}{51}\)

Answer 2

[nguyen thi van khanh]

đề bài mk chỉ cho 50 thôi ko có 51 đâu

nên mk cho bạn 1k thôi nhé