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Question Number 186187 by aba last updated on 02/Feb/23

Σ_(k=0) ^(+∞) (−1)^k (_(2k) ^n )=?

$$\underset{\mathrm{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \left(_{\mathrm{2k}} ^{\mathrm{n}} \right)=? \\ $$

Commented by mr W last updated on 02/Feb/23

for k=0 ... +∞   ((n),((2k)) ) is non−sense, no matter what  value n has. so please tell what you   mean with  ((n),((2k)) ).

$${for}\:{k}=\mathrm{0}\:...\:+\infty \\ $$$$\begin{pmatrix}{{n}}\\{\mathrm{2}{k}}\end{pmatrix}\:{is}\:{non}−{sense},\:{no}\:{matter}\:{what} \\ $$$${value}\:{n}\:{has}.\:{so}\:{please}\:{tell}\:{what}\:{you}\: \\ $$$${mean}\:{with}\:\begin{pmatrix}{{n}}\\{\mathrm{2}{k}}\end{pmatrix}. \\ $$

Commented by aba last updated on 02/Feb/23

  Σ_(k=0) ^(+∞) (−1)^k C_n ^(2k) =?

$$ \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{C}_{\mathrm{n}} ^{\mathrm{2k}} =? \\ $$

Commented by mr W last updated on 02/Feb/23

as i said for k=0 to +∞, (_n ^(2k) ) or C_n ^(2k)  is  not defined except n=0. but then the  question is Σ_(k=0) ^(+∞) (−1)^k =?, which is also  non−sense. therefore the question is  non−sense. if you have copied the  question from somewhere, check  again if you have copied correctely.  if you have created the question by  yourself, then it′s a bad attempt.

$${as}\:{i}\:{said}\:{for}\:{k}=\mathrm{0}\:{to}\:+\infty,\:\left(_{{n}} ^{\mathrm{2}{k}} \right)\:{or}\:{C}_{{n}} ^{\mathrm{2}{k}} \:{is} \\ $$$${not}\:{defined}\:{except}\:{n}=\mathrm{0}.\:{but}\:{then}\:{the} \\ $$$${question}\:{is}\:\underset{{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} =?,\:{which}\:{is}\:{also} \\ $$$${non}−{sense}.\:{therefore}\:{the}\:{question}\:{is} \\ $$$${non}−{sense}.\:{if}\:{you}\:{have}\:{copied}\:{the} \\ $$$${question}\:{from}\:{somewhere},\:{check} \\ $$$${again}\:{if}\:{you}\:{have}\:{copied}\:{correctely}. \\ $$$${if}\:{you}\:{have}\:{created}\:{the}\:{question}\:{by} \\ $$$${yourself},\:{then}\:{it}'{s}\:{a}\:{bad}\:{attempt}. \\ $$

Commented by aba last updated on 02/Feb/23

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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