Question

[Lê Hải Lan]

cho A=1/2^2+1/3^2+1/4^2+...+1/100^2. chứng minh rằng A <2

Answer 1

[Nguyễn Vũ Minh Hiếu]

Ta có :

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}\)

\(................\)

\(\frac{1}{100^2}< \frac{1}{99.100}\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1-\frac{1}{100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{99}{100}\)

Mà \(\frac{99}{100}< 2\)nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)hay \(A< 2\)

~ Hok tốt ~

Answer 2

[Taeyon]

Ta có:

A=1/2^2+1/3^2+1/4^2+...+1/100^2<1/1.2+1/2.3+1/3.4+...+1/99.100=1-1/2+1/2-1/3+...+1/99-1/100=1-1/100=99/100

mà 99/100<2 nên A<2

Áp dụng công thức: 1/n.(n+1)=1/n-1/n+1

Chúc bạn hok tốt !