Question

buna ziua, ma puteti ajuta va rog in rezolvarea acestei probleme:
$\frac{1}{2} + \frac{2}{2 {}^{2} } + \frac{3}{2 {}^{3 } } + ... + \frac{n}{ {2}^{n} } = \frac{ {2}^{n + 1} - n - 2 }{ {2}^{n} }$
unde n<Nstelat

multumesc frumos​

Answer 1

Răspuns:

Explicație pas cu pas:

$\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{n}{2^n}=\dfrac{2^{n+1}-n-2}{2^n}.~~(1)\\1.~Verificam~daca~e~adevarat~pentru~n=1:~~\dfrac{2^{1+1}-1-2}{2^1}=\dfrac{2^2-3}{2}=\dfrac{1}{2},~adevarat.\\2.~Consideram~ca~(1)~este~adevarat~pentru~n=k,~deci\\\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{k}{2^k}=\dfrac{2^{k+1}-k-2}{2^k}.~~(2)\\3.~Verificam~daca~(1)~este~adevarata~si~pentru~n=k+1,~deci$

$\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{k}{2^k}+\dfrac{k+1}{2^{k+1}} =\dfrac{2^{k+1+1}-(k+1)-2}{2^{k+1}}=\dfrac{2^{k+2}-k-3}{2^{k+1}}.~~(3)\\\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{k}{2^k}+\dfrac{k+1}{2^{k+1}} =\dfrac{2^{k+1}-k-2}{2^k}+\dfrac{k+1}{2^{k+1}}=\dfrac{2*(2^{k+1}-k-2)+k+1}{2^{k+1}}=\\=\dfrac{2*2^{k+1}-2k-4+k+1}{2^{k+1}}=\dfrac{2^{k+2}-k-3}{2^{k+1}}.~Am~demonstrat~relatia~(3),$

deci relația (1) este adevărată pentru ∀n∈N*.