# Series Math Problem: Proving a_i - n_i is a Multiple of 7

• Icebreaker
In summary, the conversation discusses the possibility of showing that a_i - n_i is a multiple of 7 for all i > 1, and a proof by induction is suggested. The conversation also involves calculations and equations to relate the values of a_i and n_i. Finally, a proof is provided that a_n is congruent to n mod 7, therefore proving that a_i - n_i is a multiple of 7.
Icebreaker
$$n_1 = 1$$

$$a_1 = 1$$

$$n_i = i$$

$$a_i = 8 a_{i-1} + 1$$

Is it possible to show that $$a_i - n_i$$ is a multiple of 7 for all $$i > 1$$?

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By my calculation, for i=2 you have a_2 = 8(a_1)+1 = 9, while n_2 = 2. So a_2 - n_2 = 7. And it just gets worse, since the a sequence is essentially geometric and the n sequence is arithmetic.

Well, here are the first few values of this series:

n_i a_i
1 1
2 9
3 73
4 585
5 4681

You can do a proof by induction. For your inductive step it will be useful to express $a_{n-1}$ in a congruence mod 7.

First, you can do something like this in paper:
ai = 8ai - 1 + 1 = 8(8ai - 2 + 1) + 1 = 82ai - 2 + 8 + 1 = 82(ai - 3 + 1) + 81 + 1 = 83ai - 3 + 82 + 81 + 1 = 84ai - 4 + 83 + 82 + 81 + 1 = 84ai - 4 + 83 + 82 + 81 + 80 (since 80 = 1) ... and finally, you can relate ai and a1.
So ai = 8i - 1ai - (i - 1) + 8i - 2 + 8i - 3 + ... + 82 + 81 + 80 = 8i - 1a1 + 8i - 2 + 8i - 3 + ... + 82 + 81 + 80 = 8i - 1 + 8i - 2 + 8i - 3 + ... + 82 + 81 + 80 (Since a1 = 1).
So you have:
$$a_i = \sum_{k = 0} ^ {i - 1} 8 ^ k$$ (1), for all i >= 1.
This part helps you get the relation between ai and i (but it's not a proof), then you can prove (1) by induction.
--------------
So $$a_i - i = \left( \sum_{k = 0} ^ {i - 1} 8 ^ k \right) - i$$.
Note that $$8 \equiv 1 \mbox{ mod } 7$$.
So what can you say about: $$\sum_{k = 0} ^ {i - 1} 8 ^ k \equiv ? \mbox{ mod } 7$$
Viet Dao,

Last edited:
I'll give it a shot.

Actually, the problem is pretty simple. What you have to prove is that
$$a_n - n\equiv 0 mod7$$
or
$$a_n\equiv n mod7$$
You know that for n=2 this is true. Now suppose
$$a_{n-1}\equiv(n-1) mod 7$$
Then
$$a_n=8a_{n-1} +1 = 7a_{n-1} +a_{n-1} +1\equiv a_{n-1} +1mod7$$
It is easily taken from there to show that this is congruent to n.

## 1. How do you prove that a_i - n_i is a multiple of 7 in a series math problem?

In order to prove that a_i - n_i is a multiple of 7, we must show that the difference between the two terms is divisible by 7. This can be done by using the properties of modular arithmetic or by using mathematical induction.

## 2. What is the significance of proving a_i - n_i is a multiple of 7 in a series math problem?

Proving that a_i - n_i is a multiple of 7 is significant because it allows us to make conclusions about the series, such as finding patterns or determining if the series is convergent or divergent.

## 3. Can you give an example of proving a_i - n_i is a multiple of 7 in a series math problem?

For example, in the series 1, 8, 15, 22, ..., we can prove that a_i - n_i (where a_i is the ith term and n_i is the nth term) is a multiple of 7 by showing that the pattern of the series is adding 7 to the previous term. Thus, a_i - n_i = 7, which is a multiple of 7.

## 4. What are some common techniques used to prove a_i - n_i is a multiple of 7 in a series math problem?

Some common techniques used to prove a_i - n_i is a multiple of 7 include using mathematical induction, modular arithmetic, or algebraic manipulation.

## 5. How is proving a_i - n_i is a multiple of 7 related to other mathematical concepts?

Proving a_i - n_i is a multiple of 7 is related to other mathematical concepts such as divisibility, modular arithmetic, and number theory. It also plays a role in solving other types of mathematical problems, such as number sequences and series.

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