do you mean finding solutions of an equation? I use graphing software and find the intersection.
i am stuck with the equation part
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pattern in a number table
- Thread starter nanase
- Start date
i am stuck with the equation part
I am a high school student who has not studied all that, and I am confident i dont require that to solve this, i can handle quadratic and cubic equations once i found how they are constructed from this tableWhat method/s have you been taught to derive the polynomial expression passing through a set of points ?
for reference look into:
en.wikipedia.org/wiki/Lagrange_polynomial ,
or Newton interpolation polynomial:
en.wikipedia.org/wiki/Newton_polynomial (and many others).
D
Deleted member 4993
Guest
Top right cell as 'x'I am reconstructing from beginning now
top right cell is n
bottom right cell is n squared
bottom left cell is n squared - n + 1
i just need to work out the sum of the diagonal from top right to bottom left in an equation and i am stuck there
Next cell (of interest) has x + (x-1) → their sum is x + [ x + (x-1)]
Next cell (of interest) has x + 2*(x-1) → their sum is x + [ x + (x-1)] + x + 2*(x-1)
Continue.....
You need to add-up 'x' cells - so that you get the sum and equate that to 5335
next cell of interest has x +x*(x-1) and their sum is x + [ x + (x-1)] + x + 2*(x-1) + .... + x + x*(x-1) ]Top right cell as 'x'
Next cell (of interest) has x + (x-1) → their sum is x + [ x + (x-1)]
Next cell (of interest) has x + 2*(x-1) → their sum is x + [ x + (x-1)] + x + 2*(x-1)
Continue.....
i can not formulate the sum, any help please
okay I graphed it using graphing software, and find the intersect to be x=22i tried using arithmetic sum formula
sum = (n/2)[2a + (n-1)d], where a is the initial value and d is the difference
in this context, a is x, and d is (x-1)
so after substitution i have 5335 = (x/2)[2x+(x-1)(x-1)]
can I solve from here?
so the bottom right number is 22^2 = 484?
D
Deleted member 4993
Guest
Correct.okay I graphed it using graphing software, and find the intersect to be x=22
so the bottom right number is 22^2 = 484?
Ah finally Thank youuuu so much master Khan, I would not know the final answer if you don't verify it.Correct.
cheers
You have a sequence
1, 5, 15, 34, 65, 111
Subtract the first from the second, the second from the third, and so on. Those are the first differences.
You get
4, 10, 19, 31, 46
No constant. Do it again to get second differences.
6, 9, 12, 15
No constant. But it is becoming regular. Do it again to get third differences.
3, 3, 3.
A constant suggests a polynomial. THIRD differences suggest a polynomial of degree three, a cubic.
The general equation for a polynomial of degree 3 is [imath]an^3 + bn^2 + cn + d.[/imath]
The general equation for a third difference of a cubic is
[math]a(n + 1)^3 + b(n + 1)^2 +c(n + 1) + d - (an^3 + bn^2 + cn + d) =\\ 3an^2 + 3an + a + 2bn + b + c =\\ n^2(3a) + n(3a +2b) + (a + b + c).[/math]
Your first first difference is for n = 1. Your second is for n = 2. Your third is for n = 3.
[math]n = 1 \implies 4 = 1^2(3a) + 1(3a + 2b) + (a + b + c) \implies \\ 4 = 7a + 3b + c \implies c = 4 - 7a - 3b.\\ n = 2 \implies 10 = 2^2(3a) + 2(3a + 2b) + a + b + 4 - 7a - 3b \implies \\ 6 = a(12 + 6 + 1 - 7) + b(4 + 1 - 3) = 12a + 2b \implies \\ b = 3 - 6a \implies 3a + 2b = 3a + 6 - 12a = 6 - 9a \text { and }\\ c = 4 - 7a - 3(3 - 6a) = 4 - 9 + 18a - 7a = 11a - 5 \implies\\ a + b + c = a + 3 - 6a + 11a - 5 = 6a - 2.\\ n = 3 \implies 19 = 3^2(3a) + 3(6 - 9a) + 6a - 2 = 27a - 27a + 6a + 18 - 2 = 6a + 16 \implies \\ 6a = 19 - 16 \implies a = \dfrac{1}{2} \implies b = 3 - 6 * \dfrac{1}{2} = 0 \text { and}\\ c = 11a - 5 = \dfrac{11}{2} - \dfrac{10}{2} = \dfrac{1}{2}. [/math]
So your equation is [imath]\dfrac{n^3}{2} + \dfrac{n}{2} + d = \dfrac{n(n^2 + 1)}{2} + d.[/imath] How do you find d?
This is high school stuff once you know about first, second, third differences.
1, 5, 15, 34, 65, 111
Subtract the first from the second, the second from the third, and so on. Those are the first differences.
You get
4, 10, 19, 31, 46
No constant. Do it again to get second differences.
6, 9, 12, 15
No constant. But it is becoming regular. Do it again to get third differences.
3, 3, 3.
A constant suggests a polynomial. THIRD differences suggest a polynomial of degree three, a cubic.
The general equation for a polynomial of degree 3 is [imath]an^3 + bn^2 + cn + d.[/imath]
The general equation for a third difference of a cubic is
[math]a(n + 1)^3 + b(n + 1)^2 +c(n + 1) + d - (an^3 + bn^2 + cn + d) =\\ 3an^2 + 3an + a + 2bn + b + c =\\ n^2(3a) + n(3a +2b) + (a + b + c).[/math]
Your first first difference is for n = 1. Your second is for n = 2. Your third is for n = 3.
[math]n = 1 \implies 4 = 1^2(3a) + 1(3a + 2b) + (a + b + c) \implies \\ 4 = 7a + 3b + c \implies c = 4 - 7a - 3b.\\ n = 2 \implies 10 = 2^2(3a) + 2(3a + 2b) + a + b + 4 - 7a - 3b \implies \\ 6 = a(12 + 6 + 1 - 7) + b(4 + 1 - 3) = 12a + 2b \implies \\ b = 3 - 6a \implies 3a + 2b = 3a + 6 - 12a = 6 - 9a \text { and }\\ c = 4 - 7a - 3(3 - 6a) = 4 - 9 + 18a - 7a = 11a - 5 \implies\\ a + b + c = a + 3 - 6a + 11a - 5 = 6a - 2.\\ n = 3 \implies 19 = 3^2(3a) + 3(6 - 9a) + 6a - 2 = 27a - 27a + 6a + 18 - 2 = 6a + 16 \implies \\ 6a = 19 - 16 \implies a = \dfrac{1}{2} \implies b = 3 - 6 * \dfrac{1}{2} = 0 \text { and}\\ c = 11a - 5 = \dfrac{11}{2} - \dfrac{10}{2} = \dfrac{1}{2}. [/math]
So your equation is [imath]\dfrac{n^3}{2} + \dfrac{n}{2} + d = \dfrac{n(n^2 + 1)}{2} + d.[/imath] How do you find d?
This is high school stuff once you know about first, second, third differences.