A = 1 + \(\frac{1}{2}\left(1+2\right)\)+ \(\frac{1}{3}\left(1+2+3\right)\)+ .... + \(\frac{1}{100}\left(1+2+3+...+100\right)\)

A = \(1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{100}\cdot\frac{100.101}{2}\)

A = \(\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)

A = \(\frac{2+3+4+...+101}{2}\)

A = \(\frac{\left(101+2\right).100}{2}\div2\)

A  = \(5150\div2=2575\)

Lời giải:

$A=1.1+2.2+3.3+...+100.100$

$=1(2-1)+2(3-1)+3(4-1)+...+100(101-1)$

$=1.2+2.3+3.4+....+100.101-(1+2+3+...+100)$

Có:

$X=1.2+2.3+3.4+....+100.101$

$3X=1.2(3-0)+2.3(4-1)+3.4(5-2)+....+100.101(102-99)$

$=3X=(1.2.3+2.3.4+3.4.5+....+100.101.102)-(0.1.2+1.2.3+...+99.100.101)$

$=100.101.102$

$\Rightarrow X=\frac{100.101.102}{3}$

$Y=1+2+3+...+100=100(100+1):2=5050$ 

$A=X-Y=\frac{100.101.102}{3}-5050=338350$

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