Thanks to the help of @maxmilgram, I've managed to figure out the solution.
SOLUTION:
- Test with n = 1, which gives:
$$0\le1\le\frac{127}{7}, \text{which is true}$$
- Assumption:
$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$
- Induction step:
$$\frac{127}{7}\cdot\ n^7\le\sum_{k=n+1}^{2n+1}k^6\le\frac{127}{7}\cdot\ (n+1)^7$$
And since:
$$\sum_{k=n+1}^{2n+1}k^6=\sum_{k=n}^{2n-1}k^6-n^6+(2n)^6+(2n+1)^6$$
we can write the induction step in terms of the assumption:
$$\frac{127}{7}n^7\le\sum_{k=n}^{2n-1}k^6-n^6+\left(2n\right)^6+\left(2n+1\right)^6\le\frac{127}{7}\left(n+7\right)^7$$
$$\frac{127}{7}n^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\le \sum _{k=n}^{2n-1}k^6\le \frac{127}{7}\left(n+7\right)^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6$$
In other words, if we could prove that:
$$\frac{127}{7}n^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\le \frac{127}{7}\left(n-1\right)^7$$ and
$$\frac{127}{7}\left(n+7\right)^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\ge \frac{127}{7}n^7$$
we would know that the induction step is indeed true. After some simplification we end up with:
$$-573n^5+395n^4-795n^3+321n^2-139n+\frac{120}{7}\le 0, \text{which is true since } n\ge 1$$
$$189n^5+395n^4+475n^3+321n^2+115n+\frac{120}{7}\ge 0, \text{which is true since } n \ge 1$$
In other words, we have shown that the induction step is true if the induction assumption also is true. Therefore, the statement is true for all positive integers n.
Best Answer
$3^{k+1}>3k^3$
We need $3^{k+1}>(k+1)^3 $
So, it sufficient to prove $3k^3>(k+1)^3\iff \left(1+\frac1k\right)^3<3$
For $k=3,\left(1+\frac1k\right)^3=\frac{64}{27}<3$
and $\left(1+\frac1{k+1}\right)^3<\left(1+\frac1k\right)^3$
$\implies \left(1+\frac1k\right)^3<3$ for $k\ge3$