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Bài 2 a:
\(A=n^3+3n^2+2n=n^3+n^2+2n^2+2n=n^2\left(n+1\right)+2n\left(n+1\right)=\left(n^2+2n\right)\left(n+1\right)=n\left(n+1\right)\left(n+2\right)\)
Mà tích 3 số nguyên liên tiếp chia hết cho 3, suy ra A chia hết cho 3
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
a,\(\frac{2}{1.3}+...\frac{2}{99.101}\)
\(=\frac{3-1}{1.3}+...+\frac{101-99}{99.101}\)
\(=\frac{3}{1.3}-\frac{1}{1.3}+...+\frac{101}{99.101}-\frac{99}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(\frac{100}{101}\)
\(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(N< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(N< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(N< 1-\frac{1}{100}\)
\(N< \frac{99}{100}< \frac{75}{100}=\frac{3}{4}\)
\(a,\)
Để A là phân số thì \(n-2\ne0\Rightarrow n\ne2\)
b, Ta có :
\(A=\frac{n+1}{n-2}=\frac{n-2+3}{n-2}=1+\frac{3}{n-2}\)
Mà \(3⋮n+2\Rightarrow n+2\inƯ(3)=\left\{\pm1;\pm3\right\}\)
Tự xét bảng
\(H=\frac{1}{a^2}+\frac{2}{a^3}+\frac{3}{a^4}+...+\frac{n}{a^{n+1}}\)
\(H=\frac{a^{n-1}+2.a^{n-2}+...+\left(n-1\right).a+n}{a^{n+1}}\)
\(H=\frac{1}{a^{n+1}}.\left[\left(a^{n-2}+a^{n-2}+a+1\right)+\left(a^{n-2}+a^{n-3}+...+a+1\right)+...+\left(a+1\right)+1\right]\)
Đặt \(Sn=1+a+a^2+...+a^n\)=>\(a.Sn=a+a^2+a^3+...+a^n+a^{n+1}\)
=> \(a.Sn-Sn=a^{n+1}-1\)=>\(Sn.\left(a-1\right)=a^{n+1}-1\)=>\(Sn=\frac{a^{n+1}-1}{a-1}\)
Khi đó \(H=\frac{1}{a^{n+1}}.\left[\frac{a^n-1}{a-1}+\frac{a^{n-1}-1}{a-1}+...+\frac{a^2-1}{a-1}+\frac{a-1}{a-1}\right]\)
\(H=\frac{1}{a^{n+1}}.\left[\frac{a^n+a^{n-1}+...+a+1-\left(n+1\right)}{a-1}\right]\)
\(H=\frac{1}{a^{n+1}}.\left[\frac{a^n+a^{n-1}+...+a+1}{a-1}-\frac{n-1}{a-1}\right]\)
\(H=\frac{1}{a^{n+1}}.\left[\frac{a^{n+1}-1}{\left(a-1\right)^2}-\frac{n-1}{a-1}\right]\)
\(H=\frac{1}{a^{n+1}}.\left[\frac{a^{n+1}}{\left(a-1\right)^2}-\frac{1}{a-1}-\frac{n+1}{a-1}\right]\)
\(H=\frac{1}{\left(a-1\right)^2}-\frac{1}{a^{n+1}.\left(a-1\right)^2}-\frac{n+1}{a^{n+1}.\left(a-1\right)}< \frac{1}{\left(a-1\right)^2}\)(đpcm)
Xong rồi đó , phù.......