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Câu 2;3;4 dễ quá... bỏ qua!!
Câu 5;6 khó quá ... khỏi làm!!
dễ quá bỏ qua!!, khó quá khỏi làm!!
cứ tiêu chí mày bạn sẽ vượt qua mọi bài toán... và nhanh chóng đạt 1đ.
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Ta có: A=2+22+23+24+25+…+260
=>A=(2+22+23)+(24+25+26)+…+(258+259+260)
=>A=(2+22+23)+23.(2+22+23)+…+257.(2+22+23)
=>A=14+23.14+…+257.14
=>A=(1+23+…+257).14
=>A=(1+23+…+257).2.7 chia hết cho 7
Vậy A chia hết cho 7
A= (2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^58+2^59+2^60)
A= 2.(1+2+2^2)+2^4.(1+2+2^3)+...+2^58(1+2+2^2)
A= 2.7+2^4.7+...+2^58.7
A= (2+2^4+...+2^58).7
=> A chia hết cho 7
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\(\text{}A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\left(2^7+2^8\right)+\left(2^9+2^{10}\right)\)
\(=2.\left(1+2\right)+2^3.\left(1+2\right)+2^5.\left(1+2\right)+2^7.\left(1+2\right)+2^9.\left(1+2\right)\)
\(=2.3+2^3.3+2^5.3+2^7.3+2^9.3\)
\(=3\left(2+2^3+2^5+2^7+2^9\right)\)
Thấy : \(2+2^3+2^5+2^7+2^9\in N\)
\(\Rightarrow3\left(2+2^3+2^5+2^7+2^9\right)⋮3\)
Hay : \(A⋮3\)( đpcm )
A = 2 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210
A = ( 2 + 22 ) + ( 23 + 24 ) + ( 25 + 26 ) + ( 27 + 28 ) + ( 29 + 210 )
A = 2.( 1 + 2 ) + 23.( 1 + 2 ) + 25.( 1 + 2 ) + 27.( 1 + 2 ) + 29.( 1 + 2 )
A = 2.3 + 23.3 + 25. 3 + 27.3 + 29.3
A = 3.( 2 + 23 + 25 + 27 + 29 ) \(⋮\)3
Vậy A chia hết cho 3.
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Ta có : \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{1599}{1600}\)
\(=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{1600}\right)\)
Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\)
\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)...\left(1-\frac{1}{1601}\right)\)
Vì \(\frac{1}{2}>\frac{1}{3};\frac{1}{4}>\frac{1}{5};\frac{1}{6}>\frac{1}{7};...;\frac{1}{1600}>\frac{1}{1601}\)
\(\Rightarrow1-\frac{1}{2}< 1-\frac{1}{3};1-\frac{1}{4}< 1-\frac{1}{5};1-\frac{1}{6}< 1-\frac{1}{7};...;1-\frac{1}{1600}< 1-\frac{1}{1601}\)
\(\Rightarrow A< B\)
hay A<\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\)
Vậy A<\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\).
a) cho A = 2+22+23+...+260
cmr A chia hết cho 3 và 7
b) cho B = 3+33+35+...+31991
cmr B chia hết cho 13
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Ta luôn có:
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{7}< \frac{6}{7}\)
\(........\)
\(\frac{1599}{1600}< \frac{1600}{1601}\)
Từ trên: \(\Rightarrow A=\frac{1}{2}.\frac{3}{4}....\frac{1599}{1600}\left(1\right)\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}...\frac{1599}{1600}< \frac{2}{3}.\frac{4}{5}....\frac{1600}{1601}\left(2\right)\)
Từ: \(\left(1\right)\left(2\right)\Rightarrow A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\left(đpcm\right)\)
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\(A=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{4033}{\left(2016.2017\right)^2}\)
\(=\dfrac{3}{1.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{4033}{2016^2.2017^2}\)
\(=\dfrac{1}{1}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{2016^2}-\dfrac{1}{2017^2}\)
\(=1-\dfrac{1}{2017^2}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
Vậy...
Ta có : A = 2 + 22 + 23 + ... + 260
A = ( 2 + 22 + 23 + 24 ) + ... + ( 257 + 258 + 259 + 260 )
A = 2( 1 + 2 + 4 + 8 ) + ... + 257( 1 + 2 + 4 + 8 )
A = 2 . 15 + ... + 257 . 15
A = 15( 2 + ... + 257 )
Vậy A ⋮ 5 và 3 vì 15 ⋮ 5 và 3
Ta có : A = 2 + 22 + 23 + ... + 260
A = ( 2 + 22 + 23 ) + ... + ( 258 + 259 + 260 )
A = 2( 1 + 2 + 3 ) + ... + 258 ( 1 + 2 + 3 )
A = 2 . 7 + ... + 258 . 7
A = 7( 2 + ... + 258 )
Vậy A ⋮ 7 vì 7 ⋮ 7
Vậy A ⋮ 5 ; 3 và 7