Question Number 8003 by Yozzia last updated on 27/Sep/16
$$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}+\mathrm{1}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{m}−{r}} \right)^{\mathrm{2}} +\left(\underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{m}−{r}} \right)^{\mathrm{2}} }\right)=? \\ $$
Commented by prakash jain last updated on 30/Sep/16
$$\underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}+\mathrm{1}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{m}−{r}} =\left(−\mathrm{1}\right)^{{m}} \underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}+\mathrm{1}} \left(−\mathrm{1}\right)^{{r}} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{m}−{r}} =\left(−\mathrm{1}\right)^{{m}} \underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}} \left(−\mathrm{1}\right)^{{r}} \\ $$$$\mathrm{Since}\:\mathrm{whole}\:\mathrm{sums}\:\mathrm{are}\:\mathrm{squared}\:\mathrm{we}\:\mathrm{can}\:\mathrm{ignore} \\ $$$$\left(−\mathrm{1}\right)^{{m}} \\ $$$${let}\:{us}\:{call}\:{C}_{{k}} =^{\mathrm{2}{m}} {C}_{{k}} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}+\mathrm{1}} \left(−\mathrm{1}\right)^{{r}} ={C}_{\mathrm{1}} −{C}_{\mathrm{3}} +{C}_{\mathrm{5}} +−…−{C}_{\mathrm{2}{m}−\mathrm{1}} \:\:…\left(\mathrm{i}\right) \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}+\mathrm{1}} \left(−\mathrm{1}\right)^{{r}} ={C}_{\mathrm{0}} −{C}_{\mathrm{2}} +{C}_{\mathrm{4}} +−…+{C}_{\mathrm{2}{m}} \:\:\:\:\:…\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} ={C}_{\mathrm{0}} +{iC}_{\mathrm{1}} −{C}_{\mathrm{2}} −{iC}_{\mathrm{3}} +−…+{C}_{\mathrm{2}{m}} \:\:\:..\left({A}\right) \\ $$$$\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} ={C}_{\mathrm{0}} −{iC}_{\mathrm{1}} −{C}_{\mathrm{2}} +{iC}_{\mathrm{3}} +−…+{C}_{\mathrm{2}{m}} \:\:\:..\left({B}\right) \\ $$$$\mathrm{from}\:\left(\mathrm{A}\right)\:\mathrm{and}\:\left(\mathrm{B}\right) \\ $$$${C}_{\mathrm{0}} −{C}_{\mathrm{2}} +{C}_{\mathrm{4}} +−..+{C}_{\mathrm{2}{m}} =\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} +\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}} \\ $$$${C}_{\mathrm{1}} −{C}_{\mathrm{3}} +{C}_{\mathrm{5}} +−…+\left(−\mathrm{1}\right)^{{m}} {C}_{\mathrm{2}{m}+\mathrm{1}} =\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} −\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}{i}} \\ $$$${substituting}\:{above}\:{values}\:{in}\:\left({i}\right)\:{amd}\:\left({ii}\left\{\right.\right. \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}+\mathrm{1}} \left(−\mathrm{1}\right)^{{r}} =\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} +\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\:^{\mathrm{2}{m}} {C}_{\mathrm{2}{r}+\mathrm{1}} \left(−\mathrm{1}\right)^{{r}} =\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} −\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}{i}} \\ $$$$\mathrm{Given}\:\mathrm{expression}\:\mathrm{in}\:\mathrm{Question}= \\ $$$$=\frac{\mathrm{1}}{\left(\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} +\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} −\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}{i}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\left(\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} +\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} −\left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} }{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} \left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{m}} } \\ $$$$\mathrm{Summation}\:\mathrm{in}\:\mathrm{comments}\:\mathrm{below} \\ $$
Commented by Yozzia last updated on 30/Sep/16
$${Great}!\:{But}\:{minor}\:{error}: \\ $$$$\left(\frac{{u}\left({i}\right)+{u}\left(−{i}\right)}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{u}\left({i}\right)−{u}\left(−{i}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{2}{u}\left({i}\right)}{\mathrm{2}}\right)\left(\frac{\mathrm{2}{u}\left(−{i}\right)}{\mathrm{2}}\right) \\ $$$$={u}\left({i}\right){u}\left(−{i}\right) \\ $$$$=\left(\mathrm{1}+{i}\right)^{\mathrm{2}{m}} \left(\mathrm{1}−{i}\right)^{\mathrm{2}{m}} =\mathrm{2}^{\mathrm{2}{m}} =\mathrm{4}^{{m}} \\ $$$$\therefore\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{m}} }=\frac{\mathrm{1}/\mathrm{4}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by prakash jain last updated on 30/Sep/16
$$\mathrm{Thanks}\:\mathrm{and}\:\mathrm{i}\:\mathrm{forgot}\:\mathrm{to}\:\mathrm{sum}\:\mathrm{as}\:\mathrm{well} \\ $$
Answered by prakash jain last updated on 30/Sep/16
$$\mathrm{see}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{comments}. \\ $$