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My math teacher said one of the problems has an answer, except the chioces are all wrong. The pattern is this: 1, 4, 9, 16, 25...

Now, I know that each is added by an odd number, and after every number, the odd number goes up to the next one. To do this, I need to know the function in terms of n (for position number), x (for coeficcient of n), y (extra term, like 3n+4, y=4), and z (the number value, like 1, 4, 9, etc.).

This is for extra credit, and I want the points.

2006-12-06 09:31:23 · 8 answers · asked by ZZ 4 in Science & Mathematics Mathematics

8 answers

n^2 for n=(1,2,3,4,5,...)

You should get to recognize perfect squares.

Here's another hint. When the differences between the numbers in a progression are equal, the equation is linear.
When the differences of the differences are equal, like this one,
then the equation is quadratic(i.e., 2nd degree). Here the differences
are 3, 5, 7, 9 so the dif of the dif is 2, 2, 2.
When the differences of the differences of the differences are equal then it's third degree, etc.

2006-12-06 09:33:35 · answer #1 · answered by albert 5 · 0 0

The function is f(x) = x^2.
They are all square numbers.

1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^5 = 25

2006-12-06 17:34:35 · answer #2 · answered by Puggy 7 · 0 0

Then 36, 49, 64, 81

all you do is take the square of each 1^2; 2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, 9^2

S= (N+1)^2 Starting with N=0 to N= infinity

2006-12-06 17:35:27 · answer #3 · answered by ۞ JønaŦhan ۞ 7 · 0 0

the function is n^2 n being element of natural numbers

2006-12-06 17:36:12 · answer #4 · answered by raj 7 · 0 0

n^2

n = 1, 2, 3, 4, or 5...

2006-12-06 17:35:31 · answer #5 · answered by asndude7 2 · 0 0

S(n) = n^2, you're looking at successive squares.

2006-12-06 17:33:34 · answer #6 · answered by Scythian1950 7 · 1 0

this question is not hard, if you notice that every term is a perfect square.

the function is n^2

2006-12-06 17:33:39 · answer #7 · answered by socialistmath 2 · 0 0

?

2006-12-06 17:39:04 · answer #8 · answered by dhcl4105 2 · 0 0

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