CMR: Q= 1/3^1+1/3^2+1/3^3+...+1/3^99+1/3^100<1/2
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CMR\(\frac{1}{3^1}-\frac{1}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}< \frac{3}{16}\)
Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(3A+A=4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4A< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt \(B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3B=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(B+3B=4B=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow B< \frac{3}{4}\) (2)
Từ (2) và (2) => \(4A< B< \frac{3}{4}\Rightarrow A< \frac{3}{16}\) (đpcm)
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\(A=\frac{7n-1}{4};B=\frac{5n+3}{12}\)
Tìm n để A,B đồng thời là các số nguyên tố
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CMR: A chia hết cho B
a) \(A=1^3+2^3+3^3+...+100^3\)
\(B=1+2+3+...+99+100\)
b) \(A=1^3+2^3+3^3+...+98^3+99^3\)
\(B=1+2+3+...+98+99\)
CMR: A chia hết cho B
a) \(A=1^3+2^3+3^3+...+99^3+100^3\)
\(B=1+2+3+...+99+100\)
b) \(A=1^3+2^3+3^3+...+98^3+99^3\)
\(B=1+2+3+...+98+99\)
CMR: 100-(\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\))=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
Có : (1+1/2+1/3+....+1/100)+(1/2+2/3+....+99/100)
= 1+(1/2+1/2)+(1/3+2/3)+.....+(1/100+99/100) ( có 99 cặp )
= 1+1+1+....+1 ( có 100 số 1 )
= 100
=> 100-(1+1/2+1/3+....+1/100)=1/2+2/3+3/4+....+99/100
Tk mk nha
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vì sao đang bằng lại chuyển thành cộng
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Vì theo quy tắc chuyển vế ta có :
a - b = c thì a = b+c
Tk mk đi
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Xem thêm câu trả lời
CMR 100-(1+\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\))= (\(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\))
Ta có :
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)\)
\(=100-1-\dfrac{1}{2}-\dfrac{1}{3}-..................-\dfrac{1}{100}\)
\(=99-\dfrac{1}{2}-\dfrac{1}{3}-................-\dfrac{1}{100}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+..................+\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+.................+\dfrac{99}{100}\)
Vậy :\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+....................+\dfrac{99}{100}\)
\(\Rightarrowđpcm\)
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CMR:
a, \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b, \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
CMR
a)\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
CMR: 100- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)
Ta có:
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(\Rightarrow100-1-\dfrac{1}{2}-...-\dfrac{1}{100}=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(\Rightarrow100=1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+...+\dfrac{1}{100}+\dfrac{99}{100}\)
\(\Rightarrow100=1+1+1+...+1\) (\(100\) số \(1\))
\(\Rightarrow100=100\)
Vậy \(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\) (Đpcm)
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CMR : \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\)\(\frac{99}{100}.\)
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