chứng minh rằng :1/2^3+1/3^3+1/4^4+...+1/29^3<5/24
minh mong cac ban giup minh nhanh chong nha!!!!!!!!!!!!
Những câu hỏi liên quan
Cho B=1/1^2+1/3^2+1/4^2+.._+1/2021.2023. Chứng minh rằng 29/62
Cho B= 1/2^2+1/3^2+1/4^2+...+1/30^2. Chứng minh rằng 29/62
Cho B= 1/2^2+1/3^2+1/4^2+...+1/30^2. Chứng minh rằng 29/62
\(B>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{30.31}\)
\(B>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{30}-\dfrac{1}{31}\)
\(B>\dfrac{1}{2}-\dfrac{1}{31}=\dfrac{29}{62}\left(đpcm\right)\)
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Cho B= 1/2^2+1/3^2+1/4^2+...+1/30^2. Chứng minh rằng 29/62
\(B>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{30.31}\)
\(B>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{30}-\dfrac{1}{31}\)
\(B>\dfrac{1}{2}-\dfrac{1}{31}=\dfrac{29}{62}\left(đpcm\right)\)
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a) Chứng minh rằng :
\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+.........+\frac{1}{29^3}
Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
Chứng minh rằng: \(\frac{29}{60}< A< \frac{2}{3}\)
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\Rightarrow A>\frac{1}{2^2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\Leftrightarrow A>\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2^2}+\frac{1}{3}-\frac{1}{10}=\frac{29}{60}\left(1\right)\)
Lại có :
\(A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(\Leftrightarrow A< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{9}=\frac{23}{36}\left(2\right)\)
Mà \(\frac{23}{36}< \frac{24}{36}=\frac{2}{3}\left(3\right)\)
Từ (1), (2) và (3) suy ra \(\frac{29}{60}< A< \frac{2}{3}\)
Câu 1
A = (x+2017).(x+2018).Chứng tỏ rằng A luôn chia hết cho2
Câu 2
Cho C=3^10+3^11+3^12+...+3^16+3^17. Chứng minh rằng C chia hết cho 40
Câu 3
D= 4^25+4^26+4^27+...=4^29+4^30. Chứng minh rằng D chia hết cho 273
Câu 2:
\(C=3^{10}+3^{11}+3^{12}+...+3^{17}.\)
\(C=\left(3^{10}+3^{11}+3^{12}+3^{13}\right)+\left(3^{14}+3^{15}+3^{16}+3^{17}\right).\)
\(C=3^{10}\left(1+3+3^2+3^3\right)+3^{14}\left(1+3+3^2+3^3\right).\)
\(C=3^{10}\left(1+3+9+27\right)+3^{14}\left(1+3+9+27\right).\)
\(C=3^{10}.40+3^{14}.40.\)
\(C=\left(3^{10}+3^{14}\right).40⋮40\left(đpcm\right).\)
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\(C=3^{10}+3^{11}+..+3^{17}\\ =\left(3^{10}+3^{11}+3^{12}+3^{13}\right)+\left(3^{14}+..+3^{17}\right)\\ =3^{10}\left(1+3+3^2+3^3\right)+3^{14}\left(1+3+3^2+3^3\right)\\ =40\left(3^{10}+3^{14}\right)⋮40\)
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1)
+Nếu x lẻ thì x+2017 là chẵn \(⋮2\)
+Nếu x là chẵn thì x+2018 cũng là chãn \(⋮2\)
\(\Rightarrow dpcm\)
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Chứng Minh Rằng
a. cho biểu thức A= 3 + 3^2+ 3^3+ 3^4+...+ 3^100 và B= 3^100-1.Chứng Minh rằng : A<B
b. Cho A= 1+4+4^2+...+4^99, B= 4^100. Chứng Minh Rằng : A<B/3
\(A=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)
\(\Leftrightarrow2A=3^{101}-3\)
\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)
\(\Leftrightarrow A< B\)
a. tính A = 3+3^2+3^3+3^4+.....+3^100
3A=3^2+3^3+3^4+3^5+....+3^100
3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100
mà B=3^100-1 => A<B
\(A=1+4+4^2+...+4^{99}\)
\(\Leftrightarrow4A=4+4^2+4^3+...+4^{100}\)
\(\Leftrightarrow3A=4^{100}-1\)
\(\Leftrightarrow A=\frac{4^{100}-1}{3}< \frac{4^{100}}{3}\)
hay A<B (đpcm)
chứng minh rằng 1/2^3 +1/3^3 +1/4^3+...+ 1/2009^3< 1/4