Question

Calculeaza:

3- \frac{3}{2} + \frac{3}{ 2^{2}} - \frac{3}{ 2^{3} } + .... + \frac{3}{ 2^{14} } - \frac{3}{ 2^{15} }

Answer 1

$S=2(1- \frac{1}{2^{14}} )=2- \frac{1}{2^{13}}$[tex]S= \frac{3}{2^{0} } - \frac{3}{2^{1}} + \frac{3}{ 2^{2}} - \frac{3}{ 2^{3} } + .... + \frac{3}{ 2^{14} } - \frac{3}{ 2^{15} }\\ S= (\frac{3}{2^{0}}+ \frac{3}{2^{2}}+...+ \frac{3}{2^{14}})-( \frac{3}{2^{1}} + \frac{3}{2^{3}} +...+ \frac{3}{2^{15}} ) \\\\ S=(3* \frac{ (\frac{1}{4})^{7} -1}{\frac{1}{4} -1} )-( \frac{3}{2}*\frac{ (\frac{1}{4})^{7} -1}{\frac{1}{4} -1} )\\\\ S= (\frac{ (\frac{1}{4})^{7} -1}{\frac{1}{4} -1})(3- \frac{3}{2} )\\\\ S= \frac{1-( \frac{1}{4} )^{7}}{ \frac{3}{4} } * \frac{3}{2}\\ [/tex]

Answer 2

3-[tex] \frac{3}{2^1}+ \frac{3}{2^2}- \frac{3}{2^3}+...+ \frac{3}{2 ^{14} }- \frac{3}{2^{15}}= 3(1-\frac{1}{2^1}+ \frac{1}{2^2}- \frac{1}{2^3}+...+ \frac{1}{2 ^{14} }- \frac{1}{2^{15}}) [/tex], paranteza este o suma a progresiei geometrice cu primul termen =1, ratia q=-1/2 si numarul de termeni n=16, conform formulei: 
$Sn=b_1* \frac{1-q^n}{1-q},avem,3S_{16}=3 \frac{1-(- \frac{1}{2})^{16} }{1-(- \frac{1}{2}) }=3* \frac{2}{3}*[1-( \frac{1}{2})^{16}]=$$2[1-( \frac{1}{2})^{16}]$