Chứng minh rằng:
\(A=\frac{1}{3^2}+\frac{1}{3^4}+......+\frac{1}{3^{4n-2}}+\frac{1}{3^{4n}}+...+\frac{1}{3^{98}}-\frac{1}{3^{100}}<0,1\)
Chứng minh rằng:
A=\(\frac{1}{3^2}+\frac{1}{3^4}+.......+\frac{1}{3^{4n-2}}+\frac{1}{3^{4n}}+....+\frac{1}{3^{98}}-\frac{1}{3^{100}}\)< 0,1
Chứng minh rằng
A= \(\frac{1}{3^{^2}}\)- \(\frac{1}{3^4}\)+.......+ \(\frac{1}{3^{4n-2}}\)- \(\frac{1}{3^{4n}}\)+.......+ \(\frac{1}{3^{98}}\)- \(\frac{1}{3^{100}}\)< 0,1
Chứng minh rằng:
a,\(\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}=\frac{5n}{3.\left(4n+3\right)}\)
b,\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}< \frac{1}{4}\)
Bài 3:
So sánh A=\(\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\frac{1}{3^8}+...+\frac{1}{3^{2n+3}}+\frac{1}{3^{4n}}+...+\frac{1}{3^{98}}-\frac{1}{3^{100}}\)với \(\frac{1}{10}\)
Chứng minh rằng:
a,\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b,\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
giúp minh với
Chứng minh rằng:
\(A=\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}< \frac{1}{4}\)
\(C=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98+1}}{3^{98}}< 100\)
\(D=\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{39}}< \frac{5}{3}\)
Bạn nào giải đúng mình tick đúng cho.Nhớ là phải giải nhanh đó
1. Chứng Minh Rằng \(\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+.....+\frac{100}{3^{100}}<\frac{3}{4}\)
2. Chứng Minh Rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2012}\)
Chứng minh rằng: \(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}<\frac{1}{50}\)???
Bạn Nào giỏi thì giúp mik với nhé? Mik đang cần gấp. thanks.