# Prove inequality consisting of sum using mathematical induction

I wish to seek help for a question from the Finnish matriculation exam in mathematics. There are multiple methods to solve the exercise, but I am ONLY interested in solving it using mathematical induction. The question is as follows:

$$\text{let}\ n\ \text{be a positive integer}, n\ge1$$ $$\text{Show that:}$$ $$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$

I have tried hard to solve this using induction, but I cannot get very far at all. Any help would be appreciated.

EDIT: What I have done so far.

1. Test with n = 1, which gives:

$$0\le1\le\frac{127}{7}, \text{which is true}$$

1. Assumption:

$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$

1. Induction step:

$$\frac{127}{7}\cdot\ n^7\le\sum_{k=n+1}^{2n+1}k^6\le\frac{127}{7}\cdot\ (n+1)^7$$

But I don't know how to prove this in terms of the assumption. I would assume that there is some way to remove the sum from this inequality. It seems to be possible to expand it using my CAS-calculator. But I just can't wrap my head around it.

I hope this enough explanation of what I have done so far.

(I might have used the wrong mathematical terms here, since English is not my first language)

• Can you show what you have tried? Where you are stuck at? Can you set up the induction step? For what values of $n$ does the induction step inequality hold? Commented Mar 20, 2021 at 15:17
• @CalvinLin I just edited my answer with what I have tried. Commented Mar 20, 2021 at 15:43
• Hint: $\sum_{k=n+1}^{2n+1}k^6=\sum_{k=n}^{2n-1}k^6-n^6+(2n)^6+(2n+1)^6$. Can you take it from here? Commented Mar 20, 2021 at 16:11
• @maxmilgram Thanks for your reply. I edited my question with what perhaps is a correct answer to the question. I would be very happy if you could look it over and tell me whether I have made some mistake anywhere. On another note, I would love to know how you came up with the statement in your comment! Commented Mar 20, 2021 at 16:46
• Yes, that's perfect. The key element in every proof by induction is to find out how to exploit the assumption. When it's about sums it is usually to split of or add some elements to the sum to make the boundaries match. In other words: make the sum from the assumption appear in the induction step and then go from there. Commented Mar 20, 2021 at 17:22

Thanks to the help of @maxmilgram, I've managed to figure out the solution.

SOLUTION:

1. Test with n = 1, which gives:

$$0\le1\le\frac{127}{7}, \text{which is true}$$

1. Assumption:

$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$

1. 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.

• You should post the solution in the answer, not as an edit to your question. Commented Mar 20, 2021 at 18:11
• @MartinR okay, I've done that now. Commented Mar 20, 2021 at 18:35
• Glad I could help! :-) Commented Mar 20, 2021 at 20:20