Question

cine stie sa ma ajute la ex 26 d)?

Answer 1

$P=(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})(1+\frac{1}{2^8})...(1+\frac{1}{2^{2^{100}}})$

Vom inmulti cu (1 - 1/2) in ambii membri, apoi vom folosi formula (a - b)(a + b) = a² - b²

[tex]P(1-\frac{1}{2})=\underline{(1-\frac{1}{2})(1+\frac{1}{2})}(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{2^{100}}})\\\\ (1-\frac{1}{2})(1+\frac{1}{2})=1^2-(\frac{1}{2})^2=1-\frac{1}{2^2}\\\\ P(1-\frac{1}{2})=\underline{(1-\frac{1}{2^2})(1+\frac{1}{2^2})}(1+\frac{1}{2^4})...(1+\frac{1}{2^{2^{100}}})\\\\ P(1-\frac{1}{2})=(1-\frac{1}{2^4})(1+\frac{1}{2^4})(1+\frac{1}{2^8})...(1+\frac{1}{2^{2^{100}}})\\\\ .\ .\ .\\\\ P(1-\frac{1}{2})=(1-\frac{1}{2^{2^{100}}})(1+\frac{1}{2^{2^{100}}})\\[/tex]
[tex]P(1-\frac{1}{2})=1-\frac{1}{2^{2^{101}}}\\ \frac{1}{2}P=1-\frac{1}{2^{2^{101}}}\\ \boxed{P=2-\frac{2}{2^{2^{101}}}}[/tex]