giup mình với

C/m nó nhỏ hơn ³/₄ hả bạn ?

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

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

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

Ta có: \(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

\(\frac{1}{4^2}<\frac{1}{3\cdot4}=\frac13-\frac14\)

...

\(\frac{1}{100^2}<\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)

Do đó: \(\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{100^2}<\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}=\frac12-\frac{1}{100}<\frac12\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{100^2}<\frac14+\frac12=\frac34\)

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

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

\(=1-\frac{1}{100}< 1\)(ĐPCM)

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

A=1+3+3^2+3^3+3^4+3^5+3^6

3A=3+3^2+3^3+3^4+3^5+3^6+3^7

3A-A=(3+3^2+3^3+3^4+3^5+3^6+3^7)-(1+3+3^2+3^3+3^4+3^5+3^6)

A=3^7-1

Vì A =3^7-1 ; B =3^7-1

=> A=B

Sửa đề:

\(A=1+3+3^2+3^3+3^4+3^5+3^6\)

\(3A=3+3^2+...+3^7\)

\(3A-A=\left(3+3^2+3^3+...+3^7\right)-\left(1+3+3^2+...+3^6\right)\)

\(2A=3^7-1\)

\(\Rightarrow A=\frac{3^7-1}{2}< 3^7-1=B\)

Vậy \(A< B\)

       S = 1 +  2 + 2² + 2³ +...+ 2⁹

      2S =       2 + 2² + 2³+...+ 2⁹ + 2¹⁰

2S - S =        2¹⁰ - 1

        S =  2¹⁰ - 1

        P = 5.2⁰ = 5 <  7 = 2³ - 1 < 2¹⁰ -1   = S
S > P  

ai nhanh mik tích cho nhé