<s>[INST] What is the sum of the digits of the sum of the digits? Problem

Let $(10^{2016}+5)^2=225N$. If $S$ is the sum of the digits of N, then find the sum of the digits of S

Attempt
Let's look at some smaller cases. We have $\dfrac{(10^{3}+5)^2}{225} = 4489$,$\dfrac{(10^{4}+5)^2}{225} = 444889$, and $\dfrac{(10^{5}+5)^2}{225} = 44448889$. Thus we see the pattern and $S = 4*2015+8*2014+9 = 24181$ and the sum of the digits of $S$ is $\boxed{16}$.
Question
Prove the result in the solution above by induction or some other method. That is, show that for $n > 1$ 
$$\dfrac{(10^{n}+5)^2}{225} =\underbrace{44\ldots4}_\text{n-1 4's}\underbrace{88\ldots8}_\text{n-2 8's}9.$$
 [/INST] hint


*

*$U_{n+1}=\frac{(10^{n+1}+5)^2}{225}=\frac{(10^{n}+5+10^n*9)^2}{225}$


= $\frac{(10^{n}+5)^2}{225}+\frac{2*(10^{n}+5)*(10^n*9)}{225}+\frac{(10^{n+1}+5)^2}{225}=U_{n}+10^{n-1}(44*10^{n-1}+4)$ after simplification
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