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Gabri18
a fost răspuns

[tex]1+ \frac{1}{1+2} + \frac{1}{1+2+3} +...+ \frac{1}{1+2+3+...+x} = \frac{400}{201} [/tex]

Răspuns :

Matepentrutoti
Stim ca 1+2+3+...+x=x(x+1)/2
1/(1+2+...+x)=2/x(x+1)=2/x-2/(x+1)
Suma se scrie:
2/1-2/2+2/2-2/3+2/3-2/4+...+2/x-2/(x+1)=400/201
2-2/(x+1)=400/201
2-400/201=2/(x+1)
2/201=2/(x+1)
x+1=201
x=200

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