Question

Ma poate ajuta cineva la d) e urgent.

Answer 1

$\frac{1}{2} + \frac{2}{2^{3}} + \frac{3}{2^{5}} +...+ \frac{n}{2^{2n-1}} = \frac{1}{9}(8- \frac{3n+4}{2^{2n-1}})$ ,n∈N*

P(1): $\frac{1}{2} = \frac{1}{9} (8- \frac{7}{2})$
$\frac{1}{2} = \frac{1}{9} *\frac{9}{2}$
$\frac{1}{2} =\frac{1}{2}$
P(n+1)
$\frac{1}{2} + \frac{2}{2^{3}} + \frac{3}{2^{5}} +...+ \frac{n}{2^{2n-1}}+ \frac{n+1}{2^{2n+1}} = \frac{1}{9}(8- \frac{3n+7}{2^{2n+1}})$
$\frac{3n+4}{2^{2n-1}}+\frac{n+1}{2^{2n+1}} = \frac{3n+7}{2^{2n+1}}$
$\frac{3n+4}{2^{2n-1}}+\frac{n+1}{2^{2n+1}} = \frac{(3n+4)(2^{2n+1})+(n+1)(2^{2n-1})}{(2^{2n+1})(2^{2n-1})}$
$\frac{(3n+4)(2^{2n+1})+(n+1)(2^{2n-1})}{(2^{2n+1})(2^{2n-1})} = \frac{3n*2^{2n}+6n+2^{3}*2^{2n}+2^{2}+n*2^{2n}+n*2^{-1}+2^{2n}+2^{-1}}{(2^{2n+1})(2^{2n-1})}$
$\frac{3n*2^{2n}+6n+2^{3}*2^{2n}+2^{2}+n*2^{2n}+n*2^{-1}+2^{2n}+2^{-1}}{(2^{2n+1})(2^{2n-1})} = \frac{(2^{2n-1})({3n+7})}{(2^{2n+1})(2^{2n-1})}$
$\frac{3n+7}{2^{2n+1}}= \frac{3n+7}{2^{2n+1}}$ ,∀ n∈ N*