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halamessi
@halamessi
January 2021
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svp bonjour demontrer que:
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slyz007
Verified answer
Bonjour
1+1/n-1/(n+1)=1+(n+1-n)/[n(n+1)]=1+1/[n(n+1)]
Donc
(1+1/n-1/(n+1))²=(1+1/[n(n+1)])²
(1+1/n-1/(n+1))²=1+2/[n(n+1)]+1/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+2n(n+1)/[n²(n+1)²]+1/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+(2n²+2n+1)/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+(n²+n²+2n+1)/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+(n²+(n+1)²)/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+n²/[n²(n+1)²]+(n+1)²/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+1/(n+1)²+1/n²
(1+1/n-1/(n+1))²=1+1/n²+1/(n+1)²
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Lista de comentários
Verified answer
Bonjour1+1/n-1/(n+1)=1+(n+1-n)/[n(n+1)]=1+1/[n(n+1)]
Donc
(1+1/n-1/(n+1))²=(1+1/[n(n+1)])²
(1+1/n-1/(n+1))²=1+2/[n(n+1)]+1/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+2n(n+1)/[n²(n+1)²]+1/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+(2n²+2n+1)/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+(n²+n²+2n+1)/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+(n²+(n+1)²)/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+n²/[n²(n+1)²]+(n+1)²/[n²(n+1)²]
(1+1/n-1/(n+1))²=1+1/(n+1)²+1/n²
(1+1/n-1/(n+1))²=1+1/n²+1/(n+1)²