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Bạn tham khảo nhé
\(a)\)Đặt \(A=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{100-1}{100}=\frac{99}{100}< 1\) ( đpcm )
Vậy \(A< 1\)
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}<1\)
\(\text{Vậy }\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
sửa đề : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)
Mk chỉ làm đc bài 2 thôi!
\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Rightarrow2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)
\(\Rightarrow2S-S=6-\frac{3}{2^9}\)
\(\Rightarrow S=6-\frac{3}{2^9}\)
Chúc bạn học tốt ( sai thì đừng ném đá ) !
Ta có :
A = \(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}\)< \(\frac{1}{1.1}+\frac{1}{1.2}+...+\frac{1}{49.50}\)
A < \(1-1+1-\frac{1}{2}+...+\frac{1}{49}-\frac{1}{50}\)
A < 1 - 1/50 = 49/50 < 2
Vậy A < 2
Bài làm
a ) \(A=\frac{9^{99}+1}{9^{100}+1}=\frac{9^{100}+1}{9^{100}+1}-\frac{9}{9^{100}+1}\)
= \(1-\frac{9}{9^{100}+1}\)
\(B=\frac{10^{98}-1}{10^{99}-1}=\frac{10^{99}-1}{10^{99}-1}-\frac{10}{10^{99}-1}\)
= \(1-\frac{10}{10^{99}-1}\)
Vì \(\frac{9}{9^{100}+1}>\frac{10}{10^{99}-1}\)
nên \(1-\frac{9}{9^{100}+1}< 1-\frac{10}{10^{99}-1}\)
\(\Rightarrow A< B\)
Bài làm
b ) \(A=\frac{5^{10}}{1+5+5^2+.....+5^9}=\frac{1+5+5^2+.....+5^9}{1+5+5^2+.....+5^9}+\frac{1+5+5^2+.....+5^8-5^9.4}{1+5+5^2+.....+5^9}\)
= \(1+\frac{1+5+5^2+.....+5^8+5^9.4}{1+5+5^2+.....+5^9}=1+5^9.3\)
\(B=\frac{6^{10}}{1+6+6^2+.....+6^9}=\frac{1+6+6^2+.....+6^9}{1+6+6^2+.....+6^9}+\frac{1+6+6^2+.....+6^8+6^9.5}{1+6+6^2+.....+6^9}\)
= \(1+\frac{1+6+6^2+.....+6^8+6^9.5}{1+6+6^2+.....+6^9}=1+6^9.4\)
Vì \(1+5^9.3< 1+6^9.4\)
nên A < B
Ta có:
\(4\left(1+5+5^2+...+5^9\right)=5\left(1+5+5^2+...+5^9\right)-\left(1+5+5^2+...+5^9\right)\)
\(=5+5^2+5^3+...+5^{10}-1-5-5^2-...-5^9\)
\(=5^{10}-1+\left(5-5\right)+\left(5^2-5^5\right)+..+\left(5^9-5^9\right)\)
\(=5^{10}-1\)
=> \(1+5+5^2+...+5^9=\frac{5^{10}-1}{4}\)
Tương tự: \(1+5+5^2+...+5^8=\frac{5^9-1}{4}\)
\(1+3+3^2+...+3^9=\frac{3^{10}-1}{2}\)
\(1+3+3^2+...+3^8=\frac{3^9-1}{2}\)
=> \(A=\frac{5^{10}-1}{5^9-1}>\frac{5^{10}-1}{5^9}=5-\frac{1}{5^9}>4;\)
\(B=\frac{3^{10}-1}{3^9-1}< \frac{3^{10}}{3^9-1}=3+\frac{3}{3^9-1}< 4;\)
=> A > B.
Câu 2 sai đề, thử rồi