Câu 1:Cho A = 1/2 + 1/3 + … + 1/100 B = 99/1 + 98/2 + … + 1/99 Tính A/B Câu 2:Tính A = 1+(1+2)+(1+2+3)+ … +(1+2+..

Giải thích các bước giải:

Câu 1:

Ta có:

$B=\dfrac{99}{1}+\dfrac{98}{2}+…+\dfrac{1}{99}$

$\to B=99+\dfrac{98}{2}+…+\dfrac{1}{99}$

$\to B=(1+\dfrac{98}{2})+(1+\dfrac{97}{3})+…+(1+\dfrac1{99})+1$

$\to B=\dfrac{2+98}{2}+\dfrac{3+97}{3}+…+\dfrac{99+1}{99}+1$

$\to B=\dfrac{100}{2}+\dfrac{100}{3}+…+\dfrac{100}{99}+1$

$\to B=100(\dfrac12+\dfrac13+…+\dfrac1{99}+\dfrac1{100})$

$\to B=100A$

$\to \dfrac{A}{B}=\dfrac{1}{100}$

Bài 2:

Ta có:

$A=\dfrac{1+(1+2)+(1+2+3)+…+(1+2+…+98)}{1.98+2.97+…+98.1}$

$\to A=\dfrac{1+\dfrac{2(2+1)}{2}+\dfrac{3(3+1)}{2}+…+\dfrac{98(98+1)}{2}}{1.98+2.97+…+98.1}$

$\to A=\dfrac{1+\dfrac{2.3}{2}+\dfrac{3.4}{2}+…+\dfrac{98.99}{2}}{1.98+2.97+…+98.1}$

$\to A=\dfrac{\dfrac12(1.2+2.3+3.4+…+98.99)}{1.98+2.97+…+98.1}$

$\to A=\dfrac12\cdot \dfrac{1.2+2.3+3.4+…+98.99}{1.98+2.97+…+98.1}$

Mà $1.2+2.3+3.4+…+98.99$

$=1.(100-98)+2.(100-97)+3.(100-96)+…+98.(100-1)$

$=1.100-1.98+2.100-2.97+3.100-3.96+…+98.100-98.1$

$=(1.100+2.100+3.100+…+98.100)-(1.98+2.97+3.96+…+98.1)$

$=100(1+2+3+..+98)-(1.98+2.97+3.96+…+98.1)$

$=100\cdot \dfrac{98(98+1)}{2}-(1.98+2.97+3.96+…+98.1)$

$=100\cdot \dfrac{98.99}{2}-(1.98+2.97+3.96+…+98.1)$

$=\dfrac{98.99.100}{2}-(1.98+2.97+3.96+…+98.1)$

Lại có:

$B=1.2+2.3+3.4+…+98.99$

$\to 3B=1.2.3+2.3.3+3.4.3+…+98.99.3$

$\to 3B=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+…+98.99.(100-97)$

$\to 3B=-0.1.2+1.2.3-1.2.3+2.3.4-2.3.4+3.4.5+…-97.98.99+98.99.100$

$\to 3B=98.99.100$

$\to B=\dfrac{98.99.100}{3}$

$\to \dfrac{98.99.100}{3}=\dfrac{98.99.100}{2}-(1.98+2.97+3.96+…+98.1)$

$\to 1.98+2.97+3.96+…+98.1=\dfrac{98.99.100}{6}$

$\to A=\dfrac12\cdot \dfrac{\dfrac{98.99.100}{3}}{\dfrac{98.99.100}{6}}$

$\to A=1$

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