Is this math problem impossible?

[InNeedOfMathsAssistance]

How do I find the second general equation for the problem 1 + 4 = 5, 2 + 5 = 12, 3 + 6 = 21 etc if the first first general equation is y=4x+x^2 where x is the first number in the sequence and y is the solution. Is there another equation that can be used that doesn’t simplify back into y=4x+x^2

 

[Steven G]

I really do not understand what you are asking as I just get totally confused when I see things like 2+5=12 and 3+6 = 21. I always thought that 2 + 10 = 12 and that 3 + 18 = 21. But you write that 2+5 = 12 and 3+6=21

Maybe you mean 1 + 4*1 = 5, 2 + 5*2 = 12, 3 + 6*3 = 21. I have no idea!

 

[InNeedOfMathsAssistance]

I really do not understand what you are asking as I just get totally confused when I see things like 2+5=12 and 3+6 = 21. I always thought that 2 + 10 = 12 and that 3 + 18 = 21. But you write that 2+5 = 12 and 3+6=21

Maybe you mean 1 + 4*1 = 5, 2 + 5*2 = 12, 3 + 6*3 = 21. I have no idea!

Yes but my question I guess is if u substitute let’s say 2 in this case for x u get x+5*2=12 and the middle value is x +3 so x+x+3*x=y then if u simplify it u get x(x+3)+x=y or x^2+4x=y now if this is one of the equations to solve the sequence what’s another equation I can use?

 

[Steven G]

Yes but my question I guess is if u substitute let’s say 2 in this case for x u get x+5*2=12 and the middle cause is x +3 so x+x+3*x=y then if u simplify it u get x(x+3)+x=y or x^2+4x=y now if this is one of the equations to solve the sequence what’s another equation I can use?

So you are saying that x + 5x = y when x=2, then for the middle you use x+3x=y. This seems strange.

Please post the problem exactly as it is written. Thank you.

 

[InNeedOfMathsAssistance]

So this is the sequence and what I meant to say is the middle value is equal to x + 3 and if u factorise u x^2+4x =y and my question is, is that the only equation for this sequence or is there another equation for it

 

[Deleted member 4993]

So this is the sequence and what I meant to say is the middle value is equal to x + 3 and if u factorise u x^2+4x =y and my question is, is that the only equation for this sequence or is there another equation for it

Did you read the response #2 - carefully?

 

[Dr.Peterson]

So this is the sequence and what I meant to say is the middle value is equal to x + 3 and if u factorise u x^2+4x =y and my question is, is that the only equation for this sequence or is there another equation for it

Please, never, ever use "u" for "you" in a question about math. It gets very confusing!

This is not really a sequence; it's looking for a function such that

f(1,4) = 5
f(2,5) = 12​

f(3,6) = 21​

You appear to have observed that in the examples of f(x, y) they always have y = x+3, so this is a very restricted sample.

There are absolutely many other possible functions that would yield these particular values. And there are also other ways to express the particular function you have observed, in addition to your observation that f(x, x+3) = x(x+4) = x^2 + 4x.

One possible function (the one suggested in post #2) is f(x, y) = x + xy = x(y + 1); I think this is likely what is expected, and when you replace y with x+3, you obtain your function.

But three points do not determine a function uniquely. I could pick any value for f(8, 11) and justify it with an equation of some sort (in particular, a cubic function).

 

[pka]

Did you read the response #2 - carefully?

View attachment 19457

I complete agree with Prof Peterson on this one.
The sequence is \(a_n=n^2+4\cdot n\)
So the entry above is incorrect \(a_4=16+16=32.\)
Look at this link. If one does go to that site to see the huge number of possibilities for any sequence.
Bookmarking this site is well worth the effort.

 

[topsquark]

So this is the sequence and what I meant to say is the middle value is equal to x + 3 and if u factorise u x^2+4x =y and my question is, is that the only equation for this sequence or is there another equation for it

I've been bored recently and I've been looking up puzzles. It is unfortunate but this kind of terminology is used a lot in these.

I particularly like
[math]\begin{cases} 1 + 5 = 4 \\ 2 + 6 = 3 \\ 3 + 7 = 10 \end{cases}[/math]
What is 5 + 1?
Answer: Depending the puzzle you get 5 + 1 = 6 (of course) or 1 + 5 = 4 so 5 + 1 = 4

-Dan

 

[pka]

I've been bored recently and I've been looking up puzzles. It is unfortunate but this kind of terminology is used a lot in these. I particularly like
[math]\begin{cases} 1 + 5 = 4 \\ 2 + 6 = 3 \\ 3 + 7 = 10 \end{cases}[/math]What is 5 + 1? Answer: Depending the puzzle you get 5 + 1 = 6 (of course) or 1 + 5 = 4 so 5 + 1 = 4

Sorry Dan that you are bored. But that is a poor excuse for being wrong.

 

[HallsofIvy]

Didn't we have another question recently that asked about the function, f(x,y) such that f(1, 4)= 6, f(2, 5)= 12, and f(3, 6)= 21?

To that post (perhaps it was on a different board) I answered that, since we have 3 equations, we have 3 pieces of information so can solve for three numbers. The simples thing to do is write f(x, y)= Ax+ By+ C for some constants, A, B, and C. Then f(1, 4)= A+ 4B+ C= 6, f(2, 5)= 2A+ 5B+ C= 12. and f(3, 6)= 3A+ 6B+ C= 21.

If we subtract the first equation from the second C cancels and we get A+ B= 6. If we subtract the first equation from the third C cancels again and we get 2A+ 2B= 15. If A+ B= 6 then 2A+ 2B= 12, not 15 so that is impossible.

Instead try f(x,y)= Ax^2+ By+ C. Then f(1, 4)= A+ 4B+ C= 6, f(2, 5)= 4A+ 5B+ C= 12, and f(3, 6)= 9A+ 6B+ C= 21. Subtract the first equation from the second again, this time getting 3A+ B= 6. Subtract the first equation from the third again getting 8A+ 2B= 15. Multiply 3A+ B= 6 by 2 to get 6A+ 2B= 12 and subtract that from 8A+ 2B= 15 to get 2A= 3 so A= 3/2. Then 3A+ B= 9/2+ B= 6 so B= 6- 9/2= 3/2 also. And A+ 4B+ C= 3/2+ 12/2+ C= 15/2+ C= 6 so C= 12/2- 15/2= -3/2.

One possible function is f(x, y)= (3/2)x^2+ (3/2)y- 3/2= (3/2)(x^2+ y- 1).

Check:
f(1, 4)= (3/2)(1+4- 1)= (3/2)(4)= 6.
f(2, 5)= (3/2)(4+ 5- 1)= (3/2)(8)= 12
f(3, 6)= (3/2)(9+ 6- 1)= (3/2)(14)= 21

So one possible function (out of many) satisfying f(1, 4)= 6, f(2, 5)= 12, and f(3, 6)= 21 is f(x, y)= (3/2)(x^2+ y- 1)