Sum of first $1+3+9+\cdots+3^n$ natural numbers. How to prove that the sum of the first $1+3+9+\cdots+3^n$ natural numbers is equal to $1^2+3^2+9^2+\cdots+(3^n)^2$? 
I've tried induction, but I can't get through the induction step. The base is simple, but in the step I can only use the induction hypothesis in a way that would give me the conclusion that the sum of the first $1+3+9+\cdots+3^{n+1}$ is equal to the sum of the first $1+3+9+\cdots+3^n$ numbers, but also the sum of the remaining $3^{n+1}$ numbers, and I don't know how to proceed from here. Any hints would be helpful.

You don't need induction; you can actually show equality with a relatively well-known (though perhaps at first "tricky") summation technique:
Well first, what is $1+3+9+\cdots+3^n$? Define that quantity to be $N$, and note that $3N = 3+9+\cdots+3^n+3^{n+1} = N + 3^{n+1}-1$, so $2N = 3^{n+1}-1$, and $N = \frac12(3^{n+1}-1)$.
Now that you know how high you are summing, you get:
$$
S = \sum_{i=1}^Ni = \frac12N(N+1) = \frac18(3^{n+1}-1)(3^{n+1}+1) = \frac18(9^{n+1}-1)
$$
Now, consider $S' = 1^2 + 3^2 + 9^2+\cdots + 3^{2n}$. Note that $9S' = 3^2 + 9^2 + \cdots + 3^{2n} + 3^{2n+2} = S' + 3^{2n+2} - 1$, so $8S' = 9^{n+1} - 1$, and $S' = \frac18(9^{n+1}-1)$.
Since $S' = S$, you are done.
