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Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có :
\(A< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\) ( đpcm )
Chúc bạn học tốt ~
cho a =1/2.3/4.5/6.....99/100 chung minh rang1/15<a<1/10
choA=1/5 mu2+1/6mu2+1/7mu2+...+1/2004mu2 chung minh rang1/65<A<1/4
chung minh rang : 1 / 2 ^ 2 + 1 / 3 ^ 2 + 1 / 4 ^ 2 + . . . + 1 / 100 ^ 2 < 99 / 100
chung minh rang 1-1/2^2-1/3^3-1/4^2-....-1/100^2>1/100
please,who can help me?
chung minh rang 1/3^2+1/4^2+1/5^2+...+1/100^2<1/2
chung minh rang 1\2 mu 2+1\3 mu 2+1\4 mu2+...+1\100 mu 2 < 1
chung minh rang :
1/3^2+1/4^2+1/5^2+1/6^2+...+1/100^2<1/2
Cho A=1/2^2+1/3^2+...+1/100^2 chung minh A<3/4
KỀM CÁCH GIẢI
chung minh rang
1/3 + 2/3^2 + ... + 100/3^100<3/4