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Sagot :

INEQUATION

Bonjour,

On développe:

1/n-2/(n+1)+ 1/(n+2)

[ (n+1)(n+2) - 2(n(n+2))+n(n+1) ]/ n(n+1)(n+2)

(n²+n+2n+2-2(n²+2n)+n²+n) / n(n+1)(n+2)

(n²+3n+2-2n²-4n+n²+n/ n(n+1)(n+2)

2 / n(n+1)(n+2)

donc 2 / n(n+1)(n+2)= 1/n-2/(n+1)+ 1/(n+2)

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