$2^2 + 4^2 + 6^2 + \dots + (2n)^2 = \dots$

The sum of the series: $(2)^2 + 2(4)^2 + 3(6)^2 + \dots$ up to $10$ terms is

The sum of the series $1+3+5^2+7+9^2+\ldots$ up to $40$ terms is equal to

$\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$ is equal to

If $0 < \theta, \phi < \frac{\pi}{2}$,$x = \sum_{n=0}^{\infty} \cos^{2n} \theta$,$y = \sum_{n=0}^{\infty} \sin^{2n} \phi$,and $z = \sum_{n=0}^{\infty} \cos^{2n} \theta \cdot \sin^{2n} \phi$,then:

$\sum_{n=1}^5 n(n^2+n+1) = $