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15 tháng 7 2019
\(a,A=1^2+3^2+5^2+...+99^2\)
\(A=1+2^2+3^2+4^2+5^2+...+99^2\)
\(A=1+2.\left(3-1\right)+3.\left(4-1\right)+...+99.\left(100-1\right)\)
\(A=\left(2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)\)
\(A=\frac{99.100.101}{3}-\frac{99.\left(99+1\right)}{2}\)
\(A=333300-4950=328350\)
14 tháng 1 2018
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
26 tháng 6 2019
a)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
=\(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)
=\(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
=\(1-\frac{1}{100!}< 1\)
\(\Rightarrow\)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
b)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
=\(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
=\(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)=\(1+1-\frac{1}{99}-\frac{1}{100}\)
=\(2-\frac{1}{99}-\frac{1}{100}< 2\)
\(\Rightarrow\)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
HP
19 tháng 12 2015
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+..+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{100!}=1-\frac{1}{100!}<1\left(đpcm\right)\)
tick nhé
5 tháng 11 2017
\(3C=1+\dfrac{2}{3}+\dfrac{2}{3^2}+...+\dfrac{100}{3^{99}}\\ \Rightarrow2C=3C-C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\\ D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\\ 2D=3D-D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)=1-\dfrac{1}{3^{99}}< 1\\ \Rightarrow D< \dfrac{1}{2}\\ \Rightarrow2C< 1+\dfrac{1}{2}\\ \RightarrowĐPCM\)
5 tháng 11 2017
2C < \(1+\dfrac{1}{2}\)
\(\Rightarrow\)C < \(\dfrac{3}{4}\)
Ta có: 2 + 22 + 23 + 24 + .... + 299 + 2100
= (2 + 22 + 23 + 24 + 25) + .... + (296 + 297 + 298 + 299 + 2100)
= 2.(1 + 2 + 22 + 23 + 24) + .... + 296.(1 + 2 + 22 + 23 + 24)
= 2 . 31 + .... + 296.31
= (2 + ... + 296) . 31 \(⋮\)31
Đặt \(D=2+2^2+2^3+2^4+2^5+...+2^{99}+2^{100}\)
\(=\left(2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)(Có 20 nhóm)
\(D=2.\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(D=2.31+2^6.31+...+2^{96}.31\)
\(D=31.\left(2+2^6+2^{96}\right)⋮31\)
Vậy ..