C/m nó nhỏ hơn ³/₄ hả bạn ?

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

                                                      \(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

                                                        \(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{3}{4}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)(ĐPCM)

A=1/4(1/1+1/2^2+...+1/50^2)

=>A=1/4+1/4*(1/2^2+...+1/50^2)

=>A<1/4+1/4*(1-1/2+1/2-1/3+...+1/49-1/50)

=>A<1/4+1/4*49/50=99/200<1/2

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(=1-\frac{1}{2020}< 1\)

giup mình với

yêu cầu đề bài là gì đây ạ?? Tính hay là CM??

tính nha

Khi n=1 thì ta sẽ có:

\(1^2+2^2+\cdots+n^2=1^2=1\)

\(\frac{n\left(n+1\right)\left(2n+1\right)}{6}=\frac{1\cdot\left(1+1\right)\left(2\cdot1+1\right)}{6}=1\)

Do đó: \(1^2+2^2+\cdots+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

Giả sử công thức đúng với n=k>=1, tức là ta sẽ có:

\(1^2+2^2+\cdots+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\) (1)

Ta sẽ cần chứng minh (1)đúng với n=k+1

\(1^2+2^2+\cdots+k^2+\left(k+1\right)^2\)

\(=\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)

\(=\frac{k\left(k+1\right)\left(2k+1\right)+6\left(k+1\right)^2}{6}=\frac{\left(k+1\right)\left(2k^2+k+6k+6\right)}{6}\)

\(=\frac{\left(k+1\right)\left(2k^2+3k+4k+6\right)}{6}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)

\(=\frac{\left(k+1\right)\left(k+1+1\right)\left\lbrack2\left(k+1\right)+1\right\rbrack}{6}\) , đúng với (1)

=>(1) luôn đúng với mọi n