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 $n &gt; 1$ 

$$\dfrac{(10^{n}+5)^2}{225} =\underbrace{44\ldots4}_\text{n-1 4's}\underbrace{88\ldots8}_\text{n-2 8's}9.$$
$$
{1\over225}(10^{n+1}+5)^2 =
{1\over9}(2\cdot10^{n}+1)^2=
{1\over9}(4\cdot10^{2n}+4\cdot10^{n}+1)=\\
{1\over9}(4(10^{2n}-1)+4(\cdot10^{n}-1)+9)=
4{10^{2n}-1\over9}+4{10^{n}-1\over9}+1=\\
4{10^{2n}-1\over10-1}+4{10^{n}-1\over10-1}+1=
4\sum_{k=0}^{2n-1}10^k+4\sum_{k=0}^{n-1}10^k+1=\\
\sum_{k=n}^{2n-1}4\cdot10^k+\sum_{k=1}^{n-1}8\cdot10^k+9=\underbrace{44\ldots4}_\text{n}\underbrace{88\ldots8}_\text{n-1}9.
$$
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