Здравствуйте!
Нужно вычислить такую сумму:

.
Вот моя попытка решения:
Пусть

-нечетное, т.е.

. Тогда получаем:
![$S_{2m+1}=(C_{2m+1}^{0})^2-(C_{2m+1}^{1})^2+(C_{2m+1}^{2})^2-(C_{2m+1}^{3})^2+...+(C_{2m+1}^{2m})^2-(C_{2m+1}^{2m+1})^2=[(C_{2m+1}^{0})^2+(C_{2m+1}^{1})^2+(C_{2m+1}^{2})^2+(C_{2m+1}^{3})^2+...+(C_{2m+1}^{2m})^2+(C_{2m+1}^{2m+1})^2]-2[(C_{2m+1}^{1})^2+(C_{2m+1}^{3})^2+...+(C_{2m+1}^{2m+1})^2]$ $S_{2m+1}=(C_{2m+1}^{0})^2-(C_{2m+1}^{1})^2+(C_{2m+1}^{2})^2-(C_{2m+1}^{3})^2+...+(C_{2m+1}^{2m})^2-(C_{2m+1}^{2m+1})^2=[(C_{2m+1}^{0})^2+(C_{2m+1}^{1})^2+(C_{2m+1}^{2})^2+(C_{2m+1}^{3})^2+...+(C_{2m+1}^{2m})^2+(C_{2m+1}^{2m+1})^2]-2[(C_{2m+1}^{1})^2+(C_{2m+1}^{3})^2+...+(C_{2m+1}^{2m+1})^2]$](https://dxdy-02.korotkov.co.uk/f/5/3/7/53736618fc15e964122d8d980fabc9fd82.png)
Так как

.
Получаем, что:
![$S_{2m+1}=C_{4m+2}^{2m+1}-2[(C_{2m+1}^{1})^2+(C_{2m+1}^{3})^2+...+(C_{2m+1}^{2m+1})^2]$ $S_{2m+1}=C_{4m+2}^{2m+1}-2[(C_{2m+1}^{1})^2+(C_{2m+1}^{3})^2+...+(C_{2m+1}^{2m+1})^2]$](https://dxdy-02.korotkov.co.uk/f/1/0/0/10089875152bfcaa9fe0d9967329584582.png)
.
Далее применяя тождество:

получаем:
![$S_{2m+1}=C_{4m+2}^{2m+1}-2[(C_{2m}^{0}+C_{2m}^{1})^2+(C_{2m}^{2}+C_{2m}^{3})^2+...+(C_{2m}^{2m-2}+C_{2m}^{2m-1})^2+(C_{2m}^{2m})^2]=C_{4m+2}^{2m+1}-2[(C_{2m}^{0})^2+(C_{2m}^{1})^2+(C_{2m}^{2})^2+(C_{2m}^{3})^2+...+(C_{2m}^{2m-1})^2+(C_{2m}^{2m})^2+2C_{2m}^{0}C_{2m}^{1}+2C_{2m}^{2}C_{2m}^{3}+...+2C_{2m}^{2m-2}C_{2m}^{2m-1}]=C_{4m+2}^{2m+1}-2[C_{4m}^{2m}+P(m)]$ $S_{2m+1}=C_{4m+2}^{2m+1}-2[(C_{2m}^{0}+C_{2m}^{1})^2+(C_{2m}^{2}+C_{2m}^{3})^2+...+(C_{2m}^{2m-2}+C_{2m}^{2m-1})^2+(C_{2m}^{2m})^2]=C_{4m+2}^{2m+1}-2[(C_{2m}^{0})^2+(C_{2m}^{1})^2+(C_{2m}^{2})^2+(C_{2m}^{3})^2+...+(C_{2m}^{2m-1})^2+(C_{2m}^{2m})^2+2C_{2m}^{0}C_{2m}^{1}+2C_{2m}^{2}C_{2m}^{3}+...+2C_{2m}^{2m-2}C_{2m}^{2m-1}]=C_{4m+2}^{2m+1}-2[C_{4m}^{2m}+P(m)]$](https://dxdy-02.korotkov.co.uk/f/d/a/f/daf4bce01d4902d2a431e9ddec826aef82.png)

.
Как уже найти

?
P.S. Если у кого-нибудь есть другое решение, предлагайте. Буду очень рад прочитать.