Can someone show me a step by step method of how to find the answer to the quadratic sequences? 10 pts!!!!?

Question

If you want me to give an example of a sequence ask!! But can you plz not use that stupid longgg formula a+(n..... i dont want to use that as our teacher didnt teach that way but ive forgotten the method she used!! thank you!

Answer 1

This looks like a GCSE question..



The best way to do it is to find the first difference, and then find the second difference. The second difference is the amount that the first difference changes by each time.



Once you have the second difference, divide it by two. This will be the coefficient (the number before) the n² term. Then you take how ever many n² you have away from each term in the sequence. So for the first term you take 1 away, the second term you take 4 away etc. Then you must calculate your new first difference - the new first difference is the coefficient of the 'n' term. Then you take away how ever many n you have from each term - you should find that everything in the sequence should now be the same. Whatever number you have left will be the number you add on at the end.



So let's take your example: 2, 5, 9, 14



First difference = 3

Second difference = 1



So we have (1/2)n² - now we'll subtract 1/2n² from each term and find a new first difference.



So we have 2 - 1/2, 5 - 2, 9 - 4.5, 14 - 8

Our new sequence is 1.5, 3, 4.5, 6

First difference is 1.5, so we have 1.5n

Take this away from everything and all the terms are 0, leaving us with the formula:



(1/2)n² + 1.5n.



Hope this helps :)

Answer 2

If you are told for sure that a sequence is quadratic, just use the general formula



y = ax^2 + bx + c



Also, look at each term in the sequence as the first, second, third, fourth, etc... the place in the sequence represents the x value, while the actual value is y. So for the sequence



2, 5, 9, 14



It would be like



(1, 2) (2, 5) (3, 9) (4, 14)



So for the equation y = ax^2 + bx + c,

Just make x and y equal the values above.

You only require three points to solve the quadratic, so all you have to do is form three equations and three unknowns.



y = ax^2 + bx + c, so for (1, 2), let x = 1 and y = 2 to get



2 = a(1)^2 + b(1) + c

2 = a + b + c



For x = 2, y = 5:

5 = a(2)^2 + b(2) + c

5 = 4a + 2b + c



For x = 3, y = 9:

9 = a(3)^2 + b(3) + c

9 = 9a + 3b + c



So your three equations three unknowns equations are:



2 = a + b + c

5 = 4a + 2b + c

9 = 9a + 3b + c



Solve this as normal.

Subtracting equation 1 by equation 2 gives us

-3 = -3a - b



Subtracting equation 2 by equation 3 gives us

-4 = -5a - b



So this is now a two equations two unknowns problem.

-3 = -3a - b

-4 = -5a - b



Subtract these two, to get a.



-3 - (-4) = (-3a) - (-5a)

1 = -3a + 5a

1 = 2a

a = 1/2



And we can get b easily, by plugging in a = 1/2 into any one of the two equations.

-4 = -5a - b, so at a = 1/2,

-4 = -5(1/2) - b

-4 = -5/2 - b

-8 = -5 - 2b

-3 = -2b

b = 3/2



Now that we have a and b, we can easily get c from the original three equations.



2 = a + b + c, a = 1/2, b = 3/2,



2 = (1/2) + (3/2) + c

2 = (4/2) + c

2 = 2 + c

0 = c



So a = 1/2, b = 3/2, c = 0, and our general equation,

y = ax^2 + bx + c

becomes



y = (1/2)x^2 + (3/2)x



So the corresponding sequence is



a(n) = (1/2)n^2 + (3/2)n

Answer 3

0 + 2 = 2

2 + 3 = 5

5 + 4 = 9

9 + 5 = 14

Then

14 + 6 = 20

20 + 7 = 27

I think you could follow the above.

Answer 4

2=2

5=2+3

9=2+3+4

14=2+3+4+5

so each of the numbers can be shown as: (1+2+3+4+5+...)-1

1+2+3+4+...+n is Sn and it is equal to n(n+1)/2

so each number of thissequence is in the form of Sn-1

=[n(n+1)/2]-1

=(n^2+n-2)/2 which n>1

Answer 5

a1=2

a2=5 ( d= a2-a1=3)

a3=9 ( d= 9-5 =4) --> d2= 4-3=1

a4= 14 (d=14-9=5) -->d2= 5-4=1 ==> d2=cte ==> an= cuadratic sequence



an= a·n²+b·n+c with



a1= a+b+c=2

a2=4a+2b+c=5 --> 3a+b=3

a3=9a+3b+c=9 --> 5a+b= 4 ---> 2a=1 --> a=1/2 ; b= 3/2; c= 0



==> an= (n²+3n)/2 = n·(n+3)/2



Saludos

Answer 6

The Richter scale is a base-10 logarithmic scale. consequently an 8.0 magnitude earthquake corresponds to ten^8 on the relative scale. consequently a million/2 magnitude could be 5x10^7. to transform back, only take the backside 10 logarithm.

Answer 7

maybe this will help =)