Question

Să se arate că pentru orice n aparține lui N au loc egalitățile:

Answer 1

Răspuns:

Explicație pas cu pas:

$(1-\frac{2}{2^2})(1-\frac{2}{3^2})(1-\frac{2}{4^2})...[1-\frac{1}{(n+1)^2}] = \frac{n+2}{2n+2}\\   \\n = 1 => 1-\frac{2}{2^2} = 1 - \frac{2}{4} = 1 - \frac{1}{2} = \frac{1}{2}\\\\P(n+1): (1-\frac{2}{2^2})(1-\frac{2}{3^2})(1-\frac{2}{4^2})...[1-\frac{1}{(n+1)^2}][1-\frac{1}{(n+2)^2}] =\frac{n+1+2}{2(n+1)+2}\\  \frac{n+2}{2n+2}\cdot [1-\frac{1}{(n+2)^2}] =\frac{n+3}{2n+4}\\\\\\ \frac{n+2}{2n+2}\cdot [\frac{(n+2)^2-1}{(n+2)^2}]=  [tex]\frac{n+2}{2n+2}\cdot\frac{n^2+4n+4-1}{(n+2)^2} =\frac{n+3}{2n+4}\\\frac{n^2+4n+3}{(2n+2)(n+2)} = \frac{n+3}{2n+4}\\\frac{(n+3)(n+1)}{2(n+1)(n+2)} =\frac{n+3}{2n+4}\\\frac{(n+3)}{2(n+2)} =\frac{n+3}{2n+4}$