Recursivity & Induction problem in number theory about two well-defined sequences.

Question

Induction Number Theory problem:
We define $ T[n] $ the $ n^{th} $ triangular number. Find $ K[n] $ as the sequence for which is true: $$ T[K[n]] = \sum_{i = 0}^{n} 9^i. $$.
Define $ K[n] $ both recursively and as a general term.
I have tried to prove that $ K[n] $ exists by induction, to find the recursive value.

Answer 1

The triangular numbers are defined as

$$T(x)= \frac{x(x+1)}{2}$$

The RHS can be expressed in closed form as

$$\sum_{i=0}^{n}9^i=\frac 18 \left(9^{n+1}-1\right)$$

So we get

$$\frac{1}{2} K(n)\Big[K(n)+1\Big]=\frac 18 \Big(9^{n+1}-1\Big)$$

and then

$$4K^2(n)+4K(n)=9^{n+1}-1$$

$$\Big[2K(n)+1\Big]^2=9^{n+1}$$

$$2K(n)+1=9^{\frac{n+1}{2}}=3^{n+1}$$

$$K(n)=\frac{3^{n+1}-1}{2}$$

This is the general term, which gives for $n=1,2,3,4...$ the sequence $$K(n)=4,13,40,121...$$

To define $K(n)$ recursively, we can note that since $2K(n)+1=3^{n+1}$, then $$2K(n+1)+1=3^{n+2}\\=3\Big[2K(n)+1\Big]$$ and so

$$K(n+1)=\frac{3[2K(n)+1]-1}{2}\\=3K(n)+1$$

with $K(0)=1$. Accordingly, $$3\cdot 1+1=4 \\ 3\cdot 4+1=13 \\ 3\cdot 13 +1=40\\ 3\cdot 40+1=121 \\ ...$$ and so on.