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

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

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

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

C/m nó nhỏ hơn ³/₄ hả bạn ?

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

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

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

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

tính nha

\(3^6:3^2+2^3.2^2-3^3.3\)

\(=3^4+2^5-3^4\)

\(=3^4-3^4+2^5\)

\(=0+2^5=2^5\)

\(3^6:3^2+2^3.2^2-3^3.3\\ =3^4+2-3^4\\ =\left(3^4-3^4\right)+2\\ =0+2\\ =2.\)

a)3¹x3²x3³x........x3¹⁰⁰

=3^{1+2+3+4+...+100}

=3^{(100+1)x(100-1+1):2}

=3^101x100:2

=3⁵⁰⁵⁰

Bài b mình không biết làm

thank nha

A = 1.2 + 2.3 + 3.4 + ...+ 59.60

3A = 1.2.3 + 2.3.3 + 3.4.3 + ...+ 59.60.3

3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) +...+ 59.60.(61-58)

3A = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ 59.60.61 - 58.59.60

3A = 58.59.60 => A = 58.59.60 : 3 = 68 440

B = 1² + 2² + 3² + 59²

B = 1.2 + 2.2 + 3.3 + 59.59

B = 2 + 4 + 9 + 3481

B = 3496

vậy A - B = 68 440 - 3 496 = 64 944

( bấm nhé )

B= bạn k ghi cái dấu ba chấm à

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

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

\(=1-\frac{1}{100}< 1\)(ĐPCM)