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Bài 1:
1) \(9A=3^3+3^5+...+3^{113}\)
\(\Rightarrow8A=9A-A=3^3+3^5+...+3^{113}-3-3^3-...-3^{111}=3^{113}-3\)
\(\Rightarrow A=\dfrac{3^{113}-3}{8}\)
2) \(9B=3^4+3^6+...+3^{202}\)
\(\Rightarrow8B=9B-B=3^4+3^6+...+3^{202}-3^2-3^4-...-3^{200}=3^{202}-3^2=3^{202}-9\)
\(\Rightarrow B=\dfrac{3^{202}-9}{8}\)
3) \(25C=5^3+5^5+...+5^{101}\)
\(\Rightarrow24C=25C-C=5^3+5^5+...+5^{101}-5-5^3-...-5^{99}=5^{101}-5\)
\(\Rightarrow C=\dfrac{5^{101}-5}{24}\)
4) \(25D=5^4+5^6+...+5^{102}\)
\(\Rightarrow24D=25D-D=5^4+5^6+...+5^{102}-5^2-5^4-...-5^{100}=5^{102}-25\)
\(\Rightarrow D=\dfrac{5^{102}-25}{24}\)
Bài 2:
a) Gọi d là UCLN(2n+1,n+1)
\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\n+1⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\2n+2⋮d\end{matrix}\right.\)
\(\Rightarrow\left(2n+2\right)-\left(2n+1\right)⋮d\Rightarrow1⋮d\)
Vậy 2n+1 và n+1 là 2 số nguyên tố cùng nhau
\(\Rightarrow\dfrac{2n+1}{n+1}\) là phân số tối giản
b) Gọi d là UCLN(2n+3,3n+4)
\(\Rightarrow\left\{{}\begin{matrix}2n+3⋮d\\3n+4⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}6n+9⋮d\\6n+8⋮d\end{matrix}\right.\)
\(\Rightarrow\left(6n+9\right)-\left(6n+8\right)⋮d\Rightarrow1⋮d\)
\(\Rightarrow\dfrac{2n+3}{3n+4}\) là phân số tối giản
\(M=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99.101}{100.100}\)
\(=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)
Xét vế phải :
\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.......+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+........+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+.......+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+........+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.......+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+........+\frac{1}{50}\left(đpcm\right)\)
\(A=1+3^2+3^4+...+3^{100}\)
\(9A=3^2+3^4+3^6+...+3^{102}\)
\(8A=3^{102}-1\)
\(\Rightarrow8A-26=3^{102}-1-26=3^{102}-27\)
Vì \(3^{102}-27⋮3\)(1)
\(3^{102}-27⋮2\)(\(3^{102}-27\)là số chẵn ) (2)
\(3^{102}-27=9\left(3^{100}-3\right)\)\(\Rightarrow3^{102}-27⋮9\)(3)
Từ (1) , (2), (3) \(\Rightarrow8A-26⋮54\)\(\left(\left(2,3,9\right)=1\right)\)
vậy ...
\(A=1+3^2+3^4+...+3^{100}\)
\(\Leftrightarrow3^2A=3^2\left(1+3^2+3^4+....+3^{100}\right)\)
\(\Leftrightarrow9A=3^2+3^4+3^6+...+3^{102}\)
\(\Leftrightarrow9A-A=\left(3^2+3^4+3^6+....+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
\(\Leftrightarrow8A=3^{102}-1\)
\(\Leftrightarrow8A-26=3^{102}-1-26=3^{102}-27\)
Ta có: \(3^{102}⋮3;27⋮3\Rightarrow3^{102}-27⋮3\left(1\right)\)
\(3^{102}-27⋮2\left(2\right)\)(3^102 -27 là số lẻ)
\(3^{102}-27=\left(3^2\right)^{51}-27=9^{51}-27⋮9\left(3\right)\)
(1)(2)(3) => 8A-26 chia hết cho 54 (đpcm)
Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học trực tuyến OLM
a: \(25S=25+5^4+...+5^{202}\)
=>24S=5^202-1
hay \(S=\dfrac{5^{202}-1}{24}\)
b:
4^30=2^30*2^30
=(2^3)^10*(2^2)^15>8^10*3^15=(8^10*3^10)*3^5>24^10*3
=>2^30+3^30+4^30>3*24^10
Ta có
\(A=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)+\left(\frac{1}{15}+\frac{1}{16}\right)\)
Vì \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}< \frac{1}{6}.3=\frac{1}{2}\)
\(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}< \frac{1}{9}.3=\frac{1}{3}\)
\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}< \frac{1}{12}.3=\frac{1}{4}\)
\(\frac{1}{15}+\frac{1}{16}< \frac{1}{10}.2=\frac{1}{5}\)
=> \(S< 2\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)< 2\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)=3\)
=> S<3 (1)
Lập luận tương tự ta có
\(S>2\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)>2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)=2\)
=> S>2 (2)
Từ (1) và (2) ta có 2 < A < 3. Vậy A không phải là số tự nhiên.
\(\dfrac{1}{25}\)S=1/54-1/56+1/58-1/510+...+1/52012-1/52014
\(\Rightarrow\)26/25.S=1/52+1/52014=1/26+...>1/26
đề có lộn không em, chị không biết giải như vậy có đúng đề không