Question

[Army of bts]

Chứng minh:

a, A = 1 + \(\frac{1}{2^2}\)+  \(\frac{1}{3^2}\)+  \(\frac{1}{4^2}\)+ ... +  \(\frac{1}{100^2}\)< 2

b, B = 1 +  \(\frac{1}{2}\)+  \(\frac{1}{3}\)+  \(\frac{1}{4}\)+ ....  +  \(\frac{1}{63}\)< 6

c, C = \(\frac{1}{2}\).  \(\frac{3}{4}\).  \(\frac{5}{6}\).   \(\frac{7}{8}\)....  \(\frac{9999}{10000}\)<  \(\frac{1}{100}\)

GIÚP MÌNH VỚI NHÉ, MÌNH ĐANG CẦN GẤP LẮM

Answer 1

[Phùng Minh Quân]

\(a)\) Ta có : 

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

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

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

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

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

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

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

\(\Rightarrow\)\(A< 1+1-\frac{1}{100}\)

\(\Rightarrow\)\(A< 2-\frac{1}{100}< 2\)

\(\Rightarrow\)\(A< 2\) ( đpcm ) 

Vậy \(A< 2\)

Chúc bạn học tốt ~ 

Answer 2

[Army of bts]

Cảm ơn bạn nhiều lắm