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Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
đặt A = 1/99 + 2/98 +3/97 +...+99/1
A = (1/99 + 1) + (2/98 +1 ) +....+ (98/2 +1) + (99/1 - 98)
A = 100/99 + 100/98 + 100/97 +...+ 100/2 + 100/100
A = 100 x (1/100 +1/99 +1/98 +...+1/2)
lại có B = 1/2 + 1/3 +.. + 1/100
=> A/B = 100
A=( 4^5/4+4^5/4^2+4^5/4^3+4^5/4^4 )+.....................+ ( 4^101/4^97+....+4^101/4^100 )
A = ( 4^4+ 4^3+4^2+4 ) + .........................................+ ( 4^4 + 4^3+4^2+4)
A= ( 4^4 + 4^ 3+ 4^2+4 ) * ( (101-5):4+1)
A = (4^4+4^3+4^2+4) * 25
A =( 256+81+16+4)*25= 8925
k cho mình nhé
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+..+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{100!}=1-\frac{1}{100!}<1\left(đpcm\right)\)
tick nhé
