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a fost răspuns

Fie :a =
[tex] \frac{1}{1 + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + ... + \frac{1}{ \sqrt{n} + \sqrt{n + 1} } [/tex]
Aflați n inclus in N daca a=20

Răspuns :

Semaka2

1/(√2+1)=(√2-1)/(√2+1)(√2-1)=√2-1)/(2-1)=√2-1)

1/(√3+√2)=(√3-√2)/(√3+√2)(√3-√2)=....=√3-√2

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1/(√n+√(n+1)=(√(n+1)-√n)/(√n+1-√n)(√(n+1)-√n)=...=√n+1)-√n

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1/(√2+1)+....+1/(√n+1)+√n)=√2-1+√3-√2+.....+√(n+1)-√n=√(n+1)-1

deci   a=√(n+1)-1=20

√(n+1)=20+1

√(n+1)=21

n+1=21²

n=21²-1