Безкрайна сума

[Gruicho]

Докажете

$\sum_{n=1}^\infty 2^{2n}\sin^4\frac a{2^n}=a^2-\sin^2a$

[KOPMOPAH]

Започваме с първия член и преобразуваме:

$~~~~2^2\sin^4\frac a2=2^2\sin^2\frac a2.\sin^2\frac a2=2^2\sin^2\frac a2(1-\cos^2\frac a2)=2^2\sin^2\frac a2-\underbrace{2^2\sin^2\frac a2\cos^2\frac a2}_{=\sin^2\left(\displaystyle 2\cdot\frac a2\right)=\sin^2a}$

Тогава сумата придобива вида:

$~~~~\sum_{n=1}^{\infty }\cdots=\cancel{2^2\sin^2\frac a2}-\sin^2a+\cancel{2^4\sin^2\frac a2^2}-\cancel{2^2\sin^2\frac a2}+ \cdots+ 2^{2n}\sin^2\frac a{2^n}-\cancel{2^{2n-2}\sin^2\frac a{2^{n-1}}}=\boxed{2^{2n}\sin^2\frac a{2^n}-\sin^2a}$

И накрая

$~~~~\lim_{n \to \infty}\left(2^{2n}\sin^2\displaystyle\frac a{2^n}-\sin^2a\right)=\lim_{n \to \infty}\left(\left(2^n\sin \displaystyle \frac a{2^n}\right)^2-\sin^2a\right)=\lim_{n \to \infty}\left(\left(\displaystyle\frac{a\sin\displaystyle \frac a{2^n}}{\displaystyle\frac a{2^n} }\right)^2-\sin^2a\right)=a^2-\sin^2a$