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

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