m-1-1-r-0-m-1-2m-2r-1-1-m-r-2-r-0-m-2m-2r-1-m-r-2-

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

\underset{m = 1}{\sum^{\infty}} (\frac{1}{{(\underset{r = 0}{\sum^{m - 1}} (\begin{matrix} 2 m \\ 2 r + 1 \end{matrix}) {(- 1)}^{m - r})}^{2} + {(\underset{r = 0}{\sum^{m}} (\begin{matrix} 2 m \\ 2 r \end{matrix}) {(- 1)}^{m - r})}^{2}}) = ?

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

\underset{r = 0}{\sum^{m - 1}} (\begin{matrix} 2 m \\ 2 r + 1 \end{matrix}) {(- 1)}^{m - r} = {(- 1)}^{m} \underset{r = 0}{\sum^{m - 1}}^{2 m} C_{2 r + 1} {(- 1)}^{r}

\underset{r = 0}{\sum^{m}} (\begin{matrix} 2 m \\ 2 r \end{matrix}) {(- 1)}^{m - r} = {(- 1)}^{m} \underset{r = 0}{\sum^{m}}^{2 m} C_{2 r} {(- 1)}^{r}

Since whole sums are squared we can ignore

{(- 1)}^{m}

l e t u s c a l l C_{k} =^{2 m} C_{k}

\underset{r = 0}{\sum^{m - 1}}^{2 m} C_{2 r + 1} {(- 1)}^{r} = C_{1} - C_{3} + C_{5} + - \dots - C_{2 m - 1} \dots (i)

\underset{r = 0}{\sum^{m}}^{2 m} C_{2 r + 1} {(- 1)}^{r} = C_{0} - C_{2} + C_{4} + - \dots + C_{2 m} \dots (ii)

{(1 + i)}^{2 m} = C_{0} + {i C}_{1} - C_{2} - {i C}_{3} + - \dots + C_{2 m} . . (A)

{(1 - i)}^{2 m} = C_{0} - {i C}_{1} - C_{2} + {i C}_{3} + - \dots + C_{2 m} . . (B)

from (A) and (B)

C_{0} - C_{2} + C_{4} + - . . + C_{2 m} = \frac{{(1 + i)}^{2 m} + {(1 - i)}^{2 m}}{2}

C_{1} - C_{3} + C_{5} + - \dots + {(- 1)}^{m} C_{2 m + 1} = \frac{{(1 + i)}^{2 m} - {(1 - i)}^{2 m}}{2 i}

s u b s t i t u t i n g a b o v e v a l u e s i n (i) a m d (i i {

\underset{r = 0}{\sum^{m - 1}}^{2 m} C_{2 r + 1} {(- 1)}^{r} = \frac{{(1 + i)}^{2 m} + {(1 - i)}^{2 m}}{2}

\underset{r = 0}{\sum^{m}}^{2 m} C_{2 r + 1} {(- 1)}^{r} = \frac{{(1 + i)}^{2 m} - {(1 - i)}^{2 m}}{2 i}

Given expression in Question =

= \frac{1}{{(\frac{{(1 + i)}^{2 m} + {(1 - i)}^{2 m}}{2})}^{2} + {(\frac{{(1 + i)}^{2 m} - {(1 - i)}^{2 m}}{2 i})}^{2}}

= \frac{1}{{(\frac{{(1 + i)}^{2 m} + {(1 - i)}^{2 m}}{2})}^{2} - {(\frac{{(1 + i)}^{2 m} - {(1 - i)}^{2 m}}{2})}^{2}}

= \frac{1}{{(1 + i)}^{2 m} {(1 - i)}^{2 m}}

= \frac{1}{2^{2 m}}

Summation in comments below

Commented by Yozzia last updated on 30/Sep/16

G r e a t! B u t m i n o r e r r o r :

{(\frac{u (i) + u (- i)}{2})}^{2} - {(\frac{u (i) - u (- i)}{2})}^{2}

= (\frac{2 u (i)}{2}) (\frac{2 u (- i)}{2})

= u (i) u (- i)

= {(1 + i)}^{2 m} {(1 - i)}^{2 m} = 2^{2 m} = 4^{m}

∴ \underset{m = 1}{\sum^{\infty}} \frac{1}{4^{m}} = \frac{1 / 4}{1 - \frac{1}{4}} = \frac{1}{3}

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

Thanks and i forgot to sum as well

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

see answer in comments .