Hello there Mathematicians!

Recently I was playing around with some numbers and I stumbled across this mathematical expression:

$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)$$

I was able to solve this equation for $a=1$:

$$\sum_{k=0}^\infty\frac{x^k}{k!}\cdot2^k=e^{2x}$$ I also managed to solve the equation for $a=2$ with the identity $\sum_{i=0}^\infty\frac{x^{2k}}{(2k)!}=\cosh(x)$:

$$\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\biggl(\sum_{l=0}^k\binom{2k}{2l}\biggr)=\mathbf[\cdots\mathbf]=\frac{1}{4}\cdot(e^{2x}+e^{-2x})+\frac{1}{2}=\frac{1}{2}\cdot(\cosh(2x)+1)$$

However, I couldn't come up with a general solution for all $a\in\Bbb{N}$. I would be very thankful if someone could either guide me towards solving this equation or post his solution here. Thank you very much in advance!

Recently I was playing around with some numbers and I stumbled across the following formal power series:

$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)$$

I was able to "simplify" the above expression for $a=1$:

$$\sum_{k=0}^\infty\frac{x^k}{k!}\cdot2^k=e^{2x}$$ I also managed to simplify the expression for $a=2$ with the identity $\sum_{i=0}^\infty\frac{x^{2k}}{(2k)!}=\cosh(x)$:

$$\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\biggl(\sum_{l=0}^k\binom{2k}{2l}\biggr)=\mathbf[\cdots\mathbf]=\frac{1}{4}\cdot(e^{2x}+e^{-2x})+\frac{1}{2}=\frac{1}{2}\cdot(\cosh(2x)+1)$$

However, I couldn't come up with a general method for all $a\in\Bbb{N}$. I would be very thankful if someone could either guide me towards simplifying this expression or post his solution here.

Hello there Mathematicians!

Recently I was playing around with some numbers and I ended up with this mathematical expression:

$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)$$

I was able to solve this equation for $a=1$:

$$\sum_{k=0}^\infty\frac{x^k}{k!}\cdot2^k=e^{2x}$$ I also managed to solve the equation for $a=2$ with the identity $\sum_{i=0}^\infty\frac{x^{2k}}{(2k)!}=\cosh(x)$:

$$\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\biggl(\sum_{l=0}^k\binom{2k}{2l}\biggr)=\mathbf[\cdots\mathbf]=\frac{1}{4}\cdot(e^{2x}+e^{-2x})+\frac{1}{2}=\frac{1}{2}\cdot(\cosh(2x)+1)$$

However, I couldn't come up with a general solution for all $a\in\Bbb{N}$. I would be very thankful if someone could either guide me towards solving this equation or post his solution here. Thank you very much in advance!

Hello there Mathematicians!

Recently I was playing around with some numbers and I stumbled across this mathematical expression:

$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)$$

I was able to solve this equation for $a=1$:

$$\sum_{k=0}^\infty\frac{x^k}{k!}\cdot2^k=e^{2x}$$ I also managed to solve the equation for $a=2$ with the identity $\sum_{i=0}^\infty\frac{x^{2k}}{(2k)!}=\cosh(x)$:

$$\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\biggl(\sum_{l=0}^k\binom{2k}{2l}\biggr)=\mathbf[\cdots\mathbf]=\frac{1}{4}\cdot(e^{2x}+e^{-2x})+\frac{1}{2}=\frac{1}{2}\cdot(\cosh(2x)+1)$$

However, I couldn't come up with a general solution for all $a\in\Bbb{N}$. I would be very thankful if someone could either guide me towards solving this equation or post his solution here. Thank you very much in advance!