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Alex Stoica
Pe: 2018-07-02T08:28:20+03:00 In: In:

UTCN 378

Cine m-ar putea ajuta la aceasta problema ?

https://imgur.com/a/zmtG33B

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  1. PhantomR
    expert (VI)

    Avem y_n=\sqrt{2}^{n+1} y_{n-1}, de unde \frac{y_n}{y_{n-1}} = \sqrt 2 ^{n+1}, n \geq 1.

    Fie k\geq 1 arbitrar. Scriind relatiile anterioare pentru n=1,2,..,k-1, k obtinem:
    \frac{y_1}{y_0}=\sqrt 2^2
    \frac{y_2}{y_1}=\sqrt 2^3
    …
    \frac{y_{k-1}}{y_{k-2}}=\sqrt 2^k
    \frac{y_k}{y_{k-1}} = \sqrt 2^{k+1}

    Inmultindu-le, ramanem cu \frac{y_k}{y_0}=\sqrt 2^{2+3+...+k+(k+1) = \sqrt 2^{\frac{(k+1)(k+2)}{2} - 1}=2^{\frac{(k+1)(k+2)-2}{4},
    de unde y_k=2^{\frac{(k+1)(k+2)+2}{4}} pentru k\geq 1.

    Obtinem atunci ca \lim_{n\to\infty} \sqrt[n^2]y_n = 2^{\lim \frac{(n+1)(n+2)+2}{4n^2}} = 2^{\frac{1}{4}}=\sqrt[4]2.

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