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Question Number 2441 by Yozzi last updated on 20/Nov/15
Prove or disprove that , for even   positive n,  2×Σ_(k=1) ^((n/2)−1) (−1)^k  ((n),(k) )+(−1)^((n/2)) ((n!)/(((n/2)!)^2 ))=−2
$${Prove}\:{or}\:{disprove}\:{that}\:,\:{for}\:{even}\: \\ $$$${positive}\:{n}, \\ $$$$\mathrm{2}×\underset{{k}=\mathrm{1}} {\overset{\frac{{n}}{\mathrm{2}}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}+\left(−\mathrm{1}\right)^{\left({n}/\mathrm{2}\right)} \frac{{n}!}{\left(\left({n}/\mathrm{2}\right)!\right)^{\mathrm{2}} }=−\mathrm{2} \\ $$
Commented by Rasheed Soomro last updated on 20/Nov/15
It is false when  n=2  In Σ_(k=1) ^((n/2)−1)   ...         (n/2)−1 should be greater than k=1  But for n=2 it is 0<1
$${It}\:{is}\:{false}\:{when}\:\:{n}=\mathrm{2} \\ $$$${In}\:\underset{{k}=\mathrm{1}} {\overset{\frac{{n}}{\mathrm{2}}−\mathrm{1}} {\sum}}\:\:… \\ $$$$\:\:\:\:\:\:\:\frac{{n}}{\mathrm{2}}−\mathrm{1}\:{should}\:{be}\:{greater}\:{than}\:{k}=\mathrm{1} \\ $$$${But}\:{for}\:{n}=\mathrm{2}\:{it}\:{is}\:\mathrm{0}<\mathrm{1} \\ $$
Commented by Yozzi last updated on 20/Nov/15
Thanks. I wanted to ensure my belief  that it is an incorrect equation  for n=2 is true.
$${Thanks}.\:{I}\:{wanted}\:{to}\:{ensure}\:{my}\:{belief} \\ $$$${that}\:{it}\:{is}\:{an}\:{incorrect}\:{equation} \\ $$$${for}\:{n}=\mathrm{2}\:{is}\:{true}. \\ $$
Commented by prakash jain last updated on 20/Nov/15
0=(1−1)^n =Σ_(i=0) ^n (−1)^i ^n C_i     = (−1)^0 ^n C_0 +Σ_(i=1) ^((n/2)−1) (−1)^i ^n C_i  + (−1)^(n/2) ^n C_(n/2)                                 + Σ_(i=(n/2)+1) ^(n−1) (−1)^i ^n C_i +(−1)^n ^n C_n    ...(1)  Given^n C_r =^n C_(n−r)  and (−1)^i =(−1)^(n−i)  (n even)    Σ_(i=1) ^((n/2)−1) (−1)^i ^n C_i  =Σ_(i=(n/2)+1) ^(n−1) (−1)^i ^n C_i       ....(2)  So (1) can be written as  0=1+2×Σ_(i=1) ^((n/2)−1) (−1)^i ^n C_i +(−1)^(n/2)  ((n!)/(((n/2)!)^2 ))+1  −2=2×Σ_(i=1) ^((n/2)−1) (−1)^i ^n C_i +(−1)^(n/2)  ((n!)/(((n/2)!)^2 ))   QED
$$\mathrm{0}=\left(\mathrm{1}−\mathrm{1}\right)^{{n}} =\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{i}} \:^{{n}} {C}_{{i}} \\ $$$$\:\:=\:\left(−\mathrm{1}\right)^{\mathrm{0}} \:^{{n}} {C}_{\mathrm{0}} +\underset{{i}=\mathrm{1}} {\overset{\frac{{n}}{\mathrm{2}}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{i}} \:^{{n}} {C}_{{i}} \:+\:\left(−\mathrm{1}\right)^{{n}/\mathrm{2}} \:^{{n}} {C}_{{n}/\mathrm{2}} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\underset{{i}=\frac{{n}}{\mathrm{2}}+\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{i}} \:^{{n}} {C}_{{i}} +\left(−\mathrm{1}\right)^{{n}} \:^{{n}} {C}_{{n}} \:\:\:…\left(\mathrm{1}\right) \\ $$$$\mathrm{Given}\:^{{n}} {C}_{{r}} =\:^{{n}} {C}_{{n}−{r}} \:{and}\:\left(−\mathrm{1}\right)^{{i}} =\left(−\mathrm{1}\right)^{{n}−{i}} \:\left({n}\:{even}\right) \\ $$$$\:\:\underset{{i}=\mathrm{1}} {\overset{\frac{{n}}{\mathrm{2}}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{i}} \:^{{n}} {C}_{{i}} \:=\underset{{i}=\frac{{n}}{\mathrm{2}}+\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{i}} \:^{{n}} {C}_{{i}} \:\:\:\:\:\:….\left(\mathrm{2}\right) \\ $$$${S}\mathrm{o}\:\left(\mathrm{1}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as} \\ $$$$\mathrm{0}=\mathrm{1}+\mathrm{2}×\underset{{i}=\mathrm{1}} {\overset{\frac{{n}}{\mathrm{2}}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{i}} \:^{{n}} {C}_{{i}} +\left(−\mathrm{1}\right)^{{n}/\mathrm{2}} \:\frac{{n}!}{\left(\left({n}/\mathrm{2}\right)!\right)^{\mathrm{2}} }+\mathrm{1} \\ $$$$−\mathrm{2}=\mathrm{2}×\underset{{i}=\mathrm{1}} {\overset{\frac{{n}}{\mathrm{2}}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{i}} \:^{{n}} {C}_{{i}} +\left(−\mathrm{1}\right)^{{n}/\mathrm{2}} \:\frac{{n}!}{\left(\left({n}/\mathrm{2}\right)!\right)^{\mathrm{2}} }\: \\ $$$$\mathrm{QED} \\ $$
Answered by prakash jain last updated on 20/Nov/15
proof given in comment.  for n=2  k=1 to (n/2)−1 results in zero terms see (2)  in comments.  The result is valid for n=2.
$${proof}\:{given}\:{in}\:{comment}. \\ $$$${for}\:{n}=\mathrm{2} \\ $$$${k}=\mathrm{1}\:{to}\:\frac{{n}}{\mathrm{2}}−\mathrm{1}\:{results}\:{in}\:{zero}\:{terms}\:{see}\:\left(\mathrm{2}\right) \\ $$$${in}\:{comments}. \\ $$$${The}\:{result}\:{is}\:{valid}\:{for}\:{n}=\mathrm{2}. \\ $$

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