Bonsoir qui peut m' aidez s'il vous-plait.... Pergunta de ideia deBac17

May 2019 1 72 Report

Bonsoir qui peut m' aidez s'il vous-plait

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Geijutsu

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Bonjour,

1. a. u1 = 1/(1²) = 1/1 = 1
u2 = (1/(1²))+(1/(2²)) = 1+(1/4) = 5/4
u3 = (1/(1²))+(1/(2²))+(1/(3²)) = (5/4)+(1/9) = (45/36)+(4/36) = 49/36
b. ∀n∈ℕ*,
$u_{n+1}-u_n=(\sum \limits_{{k=1}}^{n+1} \frac{1}{k^2})-(\sum \limits_{{k=1}}^n \frac{1}{k^2})= \frac{1}{(n+1)^2} > 0$ car (n+1)² ≥ 1 > 0
Donc $(u_n)$ est croissante.

2. a. Pour tout entier k ≥ 2,
$\frac{1}{k-1}- \frac{1}{k}=\frac{k-(k-1)}{(k-1)k}=\frac{1}{k^2-k}$
Donc $\frac{1}{k^2}\leq\frac{1}{k-1}-\frac{1}{k}\Leftrightarrow \frac{1}{k^2}\leq\frac{1}{k^2-k}\Leftrightarrow k^2}\geq k^2-k\Leftrightarrow0\geq-k\Leftrightarrow0 \leq k$ , ce qui est toujours vrai car k ≥ 2
Donc $\frac{1}{k^2}\leq\frac{1}{k-1}- \frac{1}{k}$

Je te laisse faire le reste.

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good night everybody help me please I need ideas for this topic :do you think that the virtual school could really replace the traditional school ?

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good night everybody help me please I need ideas for this topic :do you think that the virtual school could really replace the traditional school ?

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