I'll annotate the second solution, so hopefully it will make more sense.
$$5^{2(k+1)}-1 = 5^{2k+2}-1$$
The above step just expands $2(k+1) = 2k+2$ in the exponent.
$$=5^{2k+2}-5^{2k}+5^{2k}-1$$
This adds the value $5^{2k}$ and subtracts $5^{2k}$. Knowing to do this isn't obvious, but you should see that it is equivalent to the expression above.
$$=5^{2k}(25-1)+(5^{2k}-1)$$
This step did several things at once. First, it splits $5^{2k+2} = 5^{2k} \times 25$ so that we have $=5^{2k}(25) - 5^{2k} + 5^{2k} - 1$. Then we can factor out a $5^{2k}$ from $5^{2k} \times 25$ and $5^{2k} \times (-1)$ to get $5^{2k}(25-1)$. Again, it isn't obvious that this is the correct thing to do (how did we chose which of the positive or negative $5^{2k}$ to factor out?), but you should see again that it is equivalent to the statement above. After that, they just add parentheses around $5^{2k}-1$ to highlight that it is a pair (important for inductive step).
$=5^{2k}(24)+8m$
Here, we simplify the $(25-1) = 24$ (this is divisible by $8$!) and we use the fact that $5^{2k}-1 = 8m$ from our inductive statement, for some integer $m$.
$\therefore$ $5^{2(k+1)}-1$ $\Rightarrow$ $8(5^{2k}(3)+m)$
Now, we factor out the $8$. We know $5^{2k} \times 3 + m$ is an integer, so we are done.
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
Not inductive (already covered by mookid):
$3^2 \equiv 1 \pmod 8 \implies 3^{2n} \equiv (3^2)^n \equiv 1^n \equiv 1 \pmod 8$.