# k-1-2n-1-k-A-B-1-C-1-D-0-

Question Number 76813 by Rio Michael last updated on 30/Dec/19
$$\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \:=\: \\$$$$\mathrm{A}.\:\infty \\$$$$\mathrm{B}.\:\mathrm{1} \\$$$$\mathrm{C}.\:−\mathrm{1} \\$$$$\mathrm{D}.\:\mathrm{0} \\$$
Commented by JDamian last updated on 30/Dec/19
$${D} \\$$
Commented by Rio Michael last updated on 30/Dec/19
$$\mathrm{why}\:\mathrm{sir}? \\$$
Commented by JDamian last updated on 30/Dec/19
$${S}=−\mathrm{1},\:\mathrm{0},\:−\mathrm{1},\:\mathrm{0},\:\centerdot\centerdot\centerdot \\$$$${S}_{{n}} =\begin{cases}{−\mathrm{1}\:\:\:{when}\:\boldsymbol{{n}}\:{is}\:{odd}}\\{\mathrm{0}\:\:\:\:\:\:\:\:{when}\:\boldsymbol{{n}}\:{is}\:{even}\:\:}\end{cases} \\$$$${As}\:\:\mathrm{2}\boldsymbol{{n}}\:\:{is}\:{always}\:{even},\:{then}\:{S}_{{n}} =\mathrm{0} \\$$
Commented by Rio Michael last updated on 30/Dec/19
$$\mathrm{sir}\:\mathrm{k}\:=\:\mathrm{1}\:\Rightarrow\:\mathrm{S}\:=\:−\mathrm{1} \\$$$$\:\:\:\:\:\mathrm{k}\:=\:\mathrm{2}\:\Rightarrow\:\mathrm{S}\:=\:\mathrm{1} \\$$$$\mathrm{S}\:=\:−\mathrm{1},\mathrm{1},−\mathrm{1},\mathrm{1}…\:? \\$$
Commented by JDamian last updated on 30/Dec/19
$${You}\:{are}\:{confused}\:−\:{I}\:{depicted}\:\boldsymbol{{S}}_{{n}} \:{as}\:{the} \\$$$$\boldsymbol{{sum}}\:{of}\:{the}\:{squence}\:{a}_{{n}} =\:\left(−\mathrm{1}\right)^{{n}} \:{up}\:{to}\:\boldsymbol{{n}}.\:{But} \\$$$${you}\:{are}\:{using}\:\boldsymbol{{S}}_{{n}} \:{as}\:{the}\:{sequence}\:{a}_{{n}} \:{itself}. \\$$$${This}\:{problem}\:{is}\:{about}\:{the}\:{sum}\:{of}\:{the} \\$$$$\left(−\mathrm{1}\right)^{{n}} \:{sequence}. \\$$
Commented by Rio Michael last updated on 30/Dec/19
$$\mathrm{thanks}\:\mathrm{sir}. \\$$$$\\$$
Commented by abdomathmax last updated on 30/Dec/19
$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{\mathrm{2}{n}} \left(−\mathrm{1}\right)^{{k}} \:=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \left(−\mathrm{1}\right)^{{k}} \:−\mathrm{1} \\$$$$=\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}−\left(−\mathrm{1}\right)}−\mathrm{1}\:=\frac{\mathrm{2}}{\mathrm{2}}−\mathrm{1}\:=\mathrm{0} \\$$$${so}\:{the}\:{correct}\:{answer}\:{is}\:{D} \\$$
Answered by ~blr237~ last updated on 30/Dec/19
$$\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\mathrm{x}+…+\mathrm{x}^{\mathrm{n}} =\frac{\mathrm{1}−\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{1}−\mathrm{x}}\:\:\: \\$$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} =\:\left(−\mathrm{1}\right)+\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} =−\mathrm{1}+\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{2n}+\mathrm{1}} }{\mathrm{1}−\left(−\mathrm{1}\right)}=−\mathrm{1}+\mathrm{1}=\mathrm{0} \\$$