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(1+1/2)+(2+1/2)+(3+1/2)+......+(20+1/2)

Răspuns :

HyperCoder
[tex]1+2+3+...+20+20* \frac{1}{2} [/tex]
[tex]=1+2+3+...+20+10[/tex]
[tex]1+2+3+...+n = \frac{n(n+1)}{2} [/tex]
[tex]1+2+3+...+20 = \frac{20(20+1)}{2}[/tex]
[tex]1+2+3+...+20 = 210[/tex]
[tex]=>210+10=220[/tex]


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