Is it some formula to calculate $$ 1 + 2 \cdot 3 + 3 \cdot 4 \cdot 5 + 4 \cdot 5 \cdot 6 \cdot 7 + \dots + n \cdot (n+1) \cdot \dots \ (2n-1)$$
for a given $n$ without iteration?
comes from : https://stackoverflow.com/questions/34480831/recursion-sum-of-series-of-n-terms/34481682#34481682
HINT:
$$1+2\cdot3+3\cdot4\cdot5+4\cdot5\cdot6\cdot7+\dots=\frac{1}{0!}+\frac{3!}{1!}+\frac{5!}{2!}+\frac{7!}{3!}+\dots=\sum_{n=0}^{\infty}\frac{(2n+1)!}{n!}$$
Now, you can show by the limit test, that this series diverges!
For a given $m$:
$$\sum_{n=0}^{m}\frac{(2n+1)!}{n!}=\sum_{n=0}^{m}\frac{2^{2n+1}\Gamma\left(n+\frac{3}{2}\right)}{\sqrt{\pi}}=\frac{\frac{2^{3+2m}\text{E}_{\frac{5}{2}+m}\left(-\frac{1}{4}\right)\Gamma\left(\frac{5}{2}+m\right)}{\sqrt{\pi}}-\text{E}_{\frac{3}{2}}\left(-\frac{1}{4}\right)}{4\sqrt[4]{e}}$$
With $\text{E}_n(x)$ is the exponentital integral, and $\Gamma(x)$ is the gamma function.