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A = ( 1 + 2 ) + ( 22 + 23 ) + .......... + ( 210 + 211 )
= 3 + 22( 1 + 2 ) + ................ + 210( 1 + 2 )
= 3 + 22 . 3 + ........ + 210 . 3
= 3(1 + 22 + ............ 210 ) chia hết cho 3
=> ĐPCM
\(M=2+2^3+2^5+2^7+....+2^{51}\)
\(=\left(2+2^3\right)+\left(2^5+2^7\right)+....+\left(2^{49}+2^{51}\right)\)
\(=10+2^4\left(2+2^3\right)+....+2^{48}\left(2+2^3\right)\)
\(=10+2^4.10+...+2^{48}.10\)
\(=10\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮10\)
\(=2.5.\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮5\)
\(M=2+2^3+2^5+2^7+....+2^{51}.\)
\(M+2^{ }=2+2+2^3+2^5+2^7+.....+2^{51}\)
\(=\left(2+2+2^3\right)+\left(2^5+2^7+2^9\right)+....+\left(2^{47}+2^{49}+2^{51}\right)\)
\(=12+2^4\left(2+2^3+2^5\right)+......+2^{46}\left(2+2^3+2^5\right)\)
\(=12+2^4.42+....+2^{46}.42\)
\(=12+7.3.2\left(2^4+...+2^{46}\right)\)
\(\Rightarrow M=\left[12+7.3.2\left(2^4+.....+2^{46}\right)\right]-2\)
\(=10+7.3.2\left(2^4+....+2^{46}\right)\)
Ta có: \(7.3.2\left(2^4+...+2^{46}\right)⋮7\)mà 10 không chia hết cho 7
Suy M không chia hết cho 7
1/A=1.21.22.23.24.25 câu 2 làm tương tự
A.2=2.22.23.24.25.26
A.2-A=(2.22.23.24.25.2 mũ 6)-(1.21.22.23.24.25)
A=26-1
3 A=1+3+32+33+...37
3.A=3+32+33+34...+38
2A=38-1
A=(38-1):2
Câu 1 :
Ta có :
\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(A=\frac{4-1}{4}+\frac{9-1}{9}+\frac{16-1}{16}+...+\frac{10000-1}{10000}\)
\(A=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{100^2-1}{100^2}\)
\(A=\frac{2^2}{2^2}-\frac{1}{2^2}+\frac{3^2}{3^2}-\frac{1}{3^2}+\frac{4^2}{4^2}-\frac{1}{4^2}+...+\frac{100^2}{100^2}-\frac{1}{100^2}\)
\(A=1-\frac{1}{2^2}+1-\frac{1}{3^2}+1-\frac{1}{4^2}+...+1-\frac{1}{100^2}\)
\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Do từ \(2\) đến \(100\) có \(100-2+1=99\) số \(1\) nên :
\(A=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)< 99\) \(\left(1\right)\)
Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) lại có :
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B< 1-\frac{1}{100}< 1\)
\(\Rightarrow\)\(A=99-B>99-1=98\)
\(\Rightarrow\)\(A>98\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(98< A< 99\)
Vậy A không phải là số nguyên
Chúc bạn học tốt ~
Cho tổng A=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{9^2}\)+\(\frac{1}{10^2}\)
Chứng tỏ A<1
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow A< 1-\frac{1}{10}\)
\(\Rightarrow A< \frac{9}{10}\)
\(\Rightarrow A< 1\)
Vậy A<1
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
.....................
\(\frac{1}{10^2}< \frac{1}{9.10}\)
Nên : \(A=\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{9.10}\)
\(\Leftrightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{10^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\)
\(\Leftrightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{10^2}< 1-\frac{1}{10}< 1\)
30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 310 + 311
= ( 30 + 31 + 32 + 33 ) + ( 34 + 35 + 36 + 37 ) + ( 38 + 39 + 310 + 311 )
= ( 1 + 3 + 9 + 27 ) + 34 ( 1 + 3 + 9 + 27 ) + 38 ( 1 + 3 + 9 + 27 )
= 40 + 34.40 + 38.40
=40( 1 + 34 + 38 ) chia hết cho 40