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Question Number 8003 by Yozzia last updated on 27/Sep/16

Σ_(m=1) ^∞ ((1/((Σ_(r=0) ^(m−1) (((2m)),((2r+1)) ) (−1)^(m−r) )^2 +(Σ_(r=0) ^m (((2m)),((2r)) ) (−1)^(m−r) )^2 )))=?

$$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}+\mathrm{1}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{m}−{r}} \right)^{\mathrm{2}} +\left(\underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{m}−{r}} \right)^{\mathrm{2}} }\right)=? \\ $$

Commented by prakash jain last updated on 30/Sep/16

Σ_(r=0) ^(m−1) (((2m)),((2r+1)) ) (−1)^(m−r) =(−1)^m Σ_(r=0) ^(m−1) ^(2m) C_(2r+1) (−1)^r Σ_(r=0) ^m (((2m)),((2r)) ) (−1)^(m−r) =(−1)^m Σ_(r=0) ^m ^(2m) C_(2r) (−1)^r Since whole sums are squared we can ignore (−1)^m let us call C_k =^(2m) C_k Σ_(r=0) ^(m−1) ^(2m) C_(2r+1) (−1)^r =C_1 −C_3 +C_5 +−...−C_(2m−1) ...(i) Σ_(r=0) ^m ^(2m) C_(2r+1) (−1)^r =C_0 −C_2 +C_4 +−...+C_(2m) ...(ii) (1+i)^(2m) =C_0 +iC_1 −C_2 −iC_3 +−...+C_(2m) ..(A) (1−i)^(2m) =C_0 −iC_1 −C_2 +iC_3 +−...+C_(2m) ..(B) from (A) and (B) C_0 −C_2 +C_4 +−..+C_(2m) =(((1+i)^(2m) +(1−i)^(2m) )/2) C_1 −C_3 +C_5 +−...+(−1)^m C_(2m+1) =(((1+i)^(2m) −(1−i)^(2m) )/(2i)) substituting above values in (i) amd (ii{ Σ_(r=0) ^(m−1) ^(2m) C_(2r+1) (−1)^r =(((1+i)^(2m) +(1−i)^(2m) )/2) Σ_(r=0) ^m ^(2m) C_(2r+1) (−1)^r =(((1+i)^(2m) −(1−i)^(2m) )/(2i)) Given expression in Question= =(1/(((((1+i)^(2m) +(1−i)^(2m) )/2))^2 +((((1+i)^(2m) −(1−i)^(2m) )/(2i)))^2 )) =(1/(((((1+i)^(2m) +(1−i)^(2m) )/2))^2 −((((1+i)^(2m) −(1−i)^(2m) )/2))^2 )) =(1/((1+i)^(2m) (1−i)^(2m) )) =(1/2^(2m) ) Summation in comments below

$$\underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}+\mathrm{1}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{m}−{r}} =\left(−\mathrm{1}\right)^{{m}} \underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}+\mathrm{1}} \left(−\mathrm{1}\right)^{{r}} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{m}−{r}} =\left(−\mathrm{1}\right)^{{m}} \underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}} \left(−\mathrm{1}\right)^{{r}} \\ $$$$\mathrm{Since}\:\mathrm{whole}\:\mathrm{sums}\:\mathrm{are}\:\mathrm{squared}\:\mathrm{we}\:\mathrm{can}\:\mathrm{ignore} \\ $$$$\left(−\mathrm{1}\right)^{{m}} \\ $$$${let}\:{us}\:{call}\:{C}_{{k}} =^{\mathrm{2}{m}} {C}_{{k}} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}+\mathrm{1}} \left(−\mathrm{1}\right)^{{r}} ={C}_{\mathrm{1}} −{C}_{\mathrm{3}} +{C}_{\mathrm{5}} +−...−{C}_{\mathrm{2}{m}−\mathrm{1}} \:\:...\left(\mathrm{i}\right) \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}+\mathrm{1}} \left(−\mathrm{1}\right)^{{r}} ={C}_{\mathrm{0}} −{C}_{\mathrm{2}} +{C}_{\mathrm{4}} +−...+{C}_{\mathrm{2}{m}} \:\:\:\:\:...\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} ={C}_{\mathrm{0}} +{iC}_{\mathrm{1}} −{C}_{\mathrm{2}} −{iC}_{\mathrm{3}} +−...+{C}_{\mathrm{2}{m}} \:\:\:..\left({A}\right) \\ $$$$\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} ={C}_{\mathrm{0}} −{iC}_{\mathrm{1}} −{C}_{\mathrm{2}} +{iC}_{\mathrm{3}} +−...+{C}_{\mathrm{2}{m}} \:\:\:..\left({B}\right) \\ $$$$\mathrm{from}\:\left(\mathrm{A}\right)\:\mathrm{and}\:\left(\mathrm{B}\right) \\ $$$${C}_{\mathrm{0}} −{C}_{\mathrm{2}} +{C}_{\mathrm{4}} +−..+{C}_{\mathrm{2}{m}} =\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} +\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}} \\ $$$${C}_{\mathrm{1}} −{C}_{\mathrm{3}} +{C}_{\mathrm{5}} +−...+\left(−\mathrm{1}\right)^{{m}} {C}_{\mathrm{2}{m}+\mathrm{1}} =\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} −\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}{i}} \\ $$$${substituting}\:{above}\:{values}\:{in}\:\left({i}\right)\:{amd}\:\left({ii}\left\{\right.\right. \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}+\mathrm{1}} \left(−\mathrm{1}\right)^{{r}} =\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} +\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}+\mathrm{1}} \left(−\mathrm{1}\right)^{{r}} =\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} −\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}{i}} \\ $$$$\mathrm{Given}\:\mathrm{expression}\:\mathrm{in}\:\mathrm{Question}= \\ $$$$=\frac{\mathrm{1}}{\left(\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} +\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} −\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}{i}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\left(\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} +\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} −\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} \left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{m}} } \\ $$$$\mathrm{Summation}\:\mathrm{in}\:\mathrm{comments}\:\mathrm{below} \\ $$

Commented by Yozzia last updated on 30/Sep/16

Great! But minor error: (((u(i)+u(−i))/2))^2 −(((u(i)−u(−i))/2))^2 =(((2u(i))/2))(((2u(−i))/2)) =u(i)u(−i) =(1+i)^(2m) (1−i)^(2m) =2^(2m) =4^m ∴ Σ_(m=1) ^∞ (1/4^m )=((1/4)/(1−(1/4)))=(1/3)

$${Great}!\:{But}\:{minor}\:{error}: \\ $$$$\left(\frac{{u}\left({i}\right)+{u}\left(−{i}\right)}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{u}\left({i}\right)−{u}\left(−{i}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{2}{u}\left({i}\right)}{\mathrm{2}}\right)\left(\frac{\mathrm{2}{u}\left(−{i}\right)}{\mathrm{2}}\right) \\ $$$$={u}\left({i}\right){u}\left(−{i}\right) \\ $$$$=\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} \left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} =\mathrm{2}^{\mathrm{2}{m}} =\mathrm{4}^{{m}} \\ $$$$\therefore\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{m}} }=\frac{\mathrm{1}/\mathrm{4}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by prakash jain last updated on 30/Sep/16

Thanks and i forgot to sum as well

$$\mathrm{Thanks}\:\mathrm{and}\:\mathrm{i}\:\mathrm{forgot}\:\mathrm{to}\:\mathrm{sum}\:\mathrm{as}\:\mathrm{well} \\ $$

Answered by prakash jain last updated on 30/Sep/16

see answer in comments.

$$\mathrm{see}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{comments}. \\ $$

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