[RESOLVED] Pattern

[Yunie]

Consider the pattern


2^3 - 1^3 = 3 x 1^2 + 3 x 1 +1
3^2 - 2^3 = 3 x 2^2 + 3 x 2 + 1
4^3 - 3^3 = 3 x 3^2 + 3 x 3 +1
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r^3 - s^3 = 331



(a) Write down the nth line in the pattern. (Ans: (n+1)^3 - n^3 = 3n^2 + 3n +1)


(b) Find the value of r and of s in the pattern. (Ans: r = 11, s = 10)



Hey guys, please help me on this question. I really don't know anything about doing this kind of pattern question. Please help. Thanks a lot!

[obi1kenobi]

What can I help you with? Which part don't you understand? And could you please change the colour, it's difficult to read.

(a)
Solution 1:
2^3 - 1^3 = 3 * 1^2 + 3 * 1 +1
3^2 - 2^3 = 3 * 2^2 + 3 * 2 + 1
4^3 - 3^3 = 3 * 3^2 + 3 * 3 +1

Look at the numbers in bold, on the left side of the equation. Obviously, the number on the right is equal to the number on the left minus one. (1)

Since in the 1st line, the number on the left is 2, in the 2nd line it's 3....., then in the nth line it'll be n+1. (2)

From (1) and (2) we can deduce that in the n-th line, the number on the right will be n.

Now, lets take a look on the right side of the equation.

2^3 - 1^3 = 3 * 1^2 + 3 * 1 +1
3^2 - 2^3 = 3 * 2^2 + 3 * 2 + 1
4^3 - 3^3 = 3 * 3^2 + 3 * 3 +1

Both the number on the left and the one on the right change along with the row number, in the first row, they are both 1, in the second, they are 2.... so in the n-th row, they will be n.

You might ask, why are we considering these numbers, and these numbers alone. Well, the answer is simple, only these four numbers change in each equation, the others stay the same. (**)

So, we have determined that in the n-th row, the equation would be:
(n+1)^3 - n^3 = 3 x n^2 + 3 x n + 1

Solution 2:
If you know a thing or two about functions, and feel comfortable using them, try this, much simpler solution. See the part (**) and you will conclude that the left side of each equation could be written like this:

f(n) = (n+1)^3 - n^3

Similarly, you will conclude that the right side of each equation could be written like this:

g(n) = 3 * n^2 + 3 * n + 1

Since the left side and the right side are equal, we have

f(n) = g(n)

So, the n-th line would be exactly (n+1)^3 - n^3 = 3 * n^2 + 3 * n + 1.

(b)
From (1) in Solution 1, we conclude that the equality r = s + 1 is true. So, we substitute r with s + 1 in the given equality.

(s+1)^3 - s^3 = 331 ..................... (3)

Now, we can use the theorem for the difference of cubes, which states that for any two real numbers x and y, the equality x^3 - y^3 = (x-y) * (x^2+x*y+y^2) is true.

(s+1)^3 - s^3 =
= ((s+1) - s) * ((s+1)^2 + (s+1)*s + s^2) =
= 1 * (s^2 + 2*s + 1 + s^2 + s + s^2) =
= 3 * s^2 + 3*s + 1 ....................... (4)

From (3), (4)
3 * s^2 + 3*s + 1 = 331 <=>
3 * s^2 + 3*s = 330 <=>
3 * (s^2 + s) = 330 / :3 (Divide both sides by 3.) <=>
s^2 + s = 110 <=>
s * (s+1) = 110

At this point, the solution is obvious, s = 10, s + 1 = r = 11.

[Yunie]

(b)
From (1) in Solution 1, we conclude that the equality r = s + 1 is true. So, we substitute r with s + 1 in the given equality.

(s+1)^3 - s^3 = 331 ..................... (3)

Now, we can use the theorem for the difference of cubes, which states that for any two real numbers x and y, the equality x^3 - y^3 = (x-y) * (x^2+x*y+y^2) is true.

(s+1)^3 - s^3 =
= ((s+1) - s) * ((s+1)^2 + (s+1)*s + s^2) =
= 1 * (s^2 + 2*s + 1 + s^2 + s + s^2) =
= 3 * s^2 + 3*s + 1 ....................... (4)

From (3), (4)
3 * s^2 + 3*s + 1 = 331 <=>
3 * s^2 + 3*s = 330 <=>
3 * (s^2 + s) = 330 / :3 (Divide both sides by 3.) <=>
s^2 + s = 110 <=>
s * (s+1) = 110

At this point, the solution is obvious, s = 10, s + 1 = r = 11.

Thanks again for replying to my threads again. I really do appreciate your help. And, sorry for using the color pink, I will change it now. Sorry for making it difficult for you to read, my deepest apologies.

Hmm, I don't really understand about the bolded parts...Where does r = s +1 comes from? I am really confused about the whole thing in part (b)...Maybe you could somehow explain it in another way? Sorry for the trouble I have caused you, my deepest apologies again. My maths aren't that good so hope you will understand..My exams are coming so hope that I can understand and do well...Thank you again.

[obi1kenobi]

Just forget about the 331 part, and substitute r and s in each line, like so:
r^3 - s^3 = 3 * 1^2 + 3 * 1 +1
r^3 - s^3 = 3 * 2^2 + 3 * 2 +1
r^3 - s^3 = 3 * 3^2 + 3 * 3 +1

2^3 - 1^3 = 3 * 1^2 + 3 * 1 +1 ---> here the 2 would be the r and the 1 would be the s
3^2 - 2^3 = 3 * 2^2 + 3 * 2 + 1 ---> here the 3 would be the r and the 2 would be the s
4^3 - 3^3 = 3 * 3^2 + 3 * 3 +1 ---> here the 4 would be the r and the 3 would be the s

Look at the numbers in bold, on the left side of the equation. Obviously, the number to the right is equal to the number to the left minus one. ---> s = r - 1, or r = s + 1 (1)

You have caused me no trouble at all, and it's a pleasure to help.
What exactly don't you understand in part (b)?
To clarify, <=> means "is equivalent".
You know you can divide both sides of an equation with the same number (not 0) and still get a valid equation, don't you?
If you still don't understand something, tell me how old are you and I'll try to get my explanations down to your level.

[Yunie]

Now, we can use the theorem for the difference of cubes, which states that for any two real numbers x and y, the equality x^3 - y^3 = (x-y) * (x^2+x*y+y^2) is true.

(s+1)^3 - s^3 =
= ((s+1) - s) * ((s+1)^2 + (s+1)*s + s^2) =
= 1 * (s^2 + 2*s + 1 + s^2 + s + s^2) =
= 3 * s^2 + 3*s + 1 ....................... (4)

From (3), (4)
3 * s^2 + 3*s + 1 = 331 <=>
3 * s^2 + 3*s = 330 <=>
3 * (s^2 + s) = 330 / :3 (Divide both sides by 3.) <=>
s^2 + s = 110 <=>
s * (s+1) = 110

At this point, the solution is obvious, s = 10, s + 1 = r = 11.

I still don't understand the bolded parts...Sorry again...By the way, I am 16 this year.

[obi1kenobi]

Ok, look at the equation x^3 - y^3 = (x-y) * (x^2+x*y+y^2). By multiplying (x-y) * (x^2+x*y+y^2) = x^3 + x^2 * y + x * y^2 - x^2 * y - x * y^2 - y^3 = x^3 - y^3. (Because both the bolded and the underlined expressions cancel each other out.)

Now, if we substitute x with (s+1) and y with s, we get this

(s+1)^3 - s^3 = ((s+1) - s) * ((s+1)^2 + (s+1)*s + s^2)

In the next line, ((s+1) - s) = (s + 1 - s) = 1, so that's where the one comes from, and the expression (s^2 + 2*s + 1 + s^2 + s + s^2) is equivalent to ((s+1)^2 + (s+1)*s + s^2). See? The bolded parts are equivalent (square of a binomial), and so are the underlined parts, when you get rid of the parentheses.

We can sum up all the bolded expressions. The same goes for the underlined ones.
(s^2 + 2*s + 1 + s^2 + s + s^2) =
= 3 * s^2 + 3 * s + 1

Since (s+1)^3 - s^3 = 331 and (s+1)^3 - s^3 = 3 * s^2 + 3*s + 1, then 3 * s^2 + 3*s + 1 = 331.

So, if we subtract 1 from both sides, we get 3 * s^2 + 3*s + 1 - 1 = 331 - 1, or 3 * s^2 + 3*s = 330.

Now, divide both sides by 3, and the result is this s^2 + s = 110.

That is one and the same with this equality: s * s + s * 1 = 110.

s occurs in both s * s and s * 1, so we can put it outside the parenthesis, and we get s * (s + 1) = 110.

Now, which two consecutive numbers, when multiplied, give a result of 110? Well 10 and 11, of course! Since s < s + 1 (obviously ), and 10 < 11, then it must be s = 10 and s + 1 = 11.

We proved that r = s + 1 and that s + 1 = 11, and the conclusion from these two equalities is that r = 11.

Btw, I'll be away for about a week, so when I get back I'll check whether you have any more questions or not.

[Yunie]

I finally understand the question and know how to do it...Thank you obi1kenobi so much!