Finding the sum of the digits

We note that:

$\begin{array} {ccc} \small \underline{\text{Number of digits}} & & \small \underline{\text{Sum of digits}} \\ 1 & 1^2 = 1 & s_1 = 1 = 1^2 \\ 2 & 11^2 = 121 & s_2 = 4 = 2^2 \\ 3 & 111^2 = 12321 & s_3 = 9 = 3^2 \\ ... & ... = ... & ... = ... = ... \\ n & \underbrace{111...111^2}_{\text{\# of 1s = n}} = \overline{123...n...321} & s_n = n^2 \end{array}$

Let us prove the claim that $s_n = n^2$ by induction for all $n$ . For convenience, let $a_n = \underbrace{111...111}_{\text{\# of 1s = n}}$ and $s(k)$ be the function to find sum of digits of integer $k$ .

Proof:

• For $n=1$ , $s_1 = 1 = 1^2$ , the claim is true.

• Assuming the claim is true for $n$ , then

$\begin{aligned} a_{\color{#3D99F6}n+1} & = 10^n + a_n \\ a_{\color{#3D99F6}n+1}^2 & = \left(10^n + a_n\right)^2 \\ & = 10^{2n} + 2\left(10^n \right)a_n + a_n^2 \\ \implies s_{\color{#3D99F6}n+1} & = s\left(10^{2n} + 2\left(10^n \right)a_n + a_n^2\right) \\ & = s\left(10^{2n}\right) + s\left(2\left(10^n \right)a_n\right) + s\left(a_n^2\right) \\ & = 1 + 2s\left(a_n\right) + s_n \\ & = 1 + 2n + n^2 \\ & = (n+1)^2 \end{aligned}$

$\quad$ Therefore, the claim is also true for $n+1$ and hence true for all $n \ge 1$ .

For $n=6$ , we have $s_6 = 6^2 = \boxed{36}$ .