Question 414398: I have no idea what the pattern is for: 10,8,3,0,4,20,53,__,__,__
Can anyone help? Thank you.
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website!
List the terms vertically
a1 = 10
a2 = 8
a3 = 3
a4 = 0
a5 = 4
a6 = 20
a7 = 53
Form the next column by subtracting each number from the number
just below it, and writing this difference beside the number.
This is called the list of first differences.
a1 = 10 -2
a2 = 8 -5
a3 = 3 -3
a4 = 0 4
a5 = 4 16
a6 = 20 33
a7 = 53
Form the next column by subtracting each number from the number
just below it, and writing this difference beside the number.
This is called the list of second differences.
a1 = 10 -2 -3
a2 = 8 -5 2
a3 = 3 -3 7
a4 = 0 4 12
a5 = 4 16 17
a6 = 20 33
a7 = 53
Form the next column by subtracting each number from the number
just below it, and writing this difference beside the number.
This is called the list of third differences.
a1 = 10 -2 -3 5
a2 = 8 -5 2 5
a3 = 3 -3 7 5
a4 = 0 4 12 5
a5 = 4 16 17
a6 = 20 33
a7 = 53
The third differences are all the same. So we stop finding
differences for we now know to assume a third degree polynomial,
the same as the number of differences required to achieve a
column with all the same number.
an = An³ + Bn² + Cn + D
There are 4 unknown coefficients so we use the n = 1,2,3,4
Substituting n = 1
a1 = A(1)³ + B(1)² + C(1) + D = 10
a1 = A + B + C + D = 10
Substituting n = 2
a2 = A(2)³ + B(2)² + C(2) + D = 8
a2 = A(8) + B(4) + C(2) + D = 8
a2 = 8A + 4B + 2C + D = 8
Substituting n = 3
a3 = A(3)³ + B(3)² + C(3) + D = 3
a3 = A(27) + B(9) + C(3) + D = 3
a3 = 27A + 9B + 3C + D = 3
Substituting n = 4
a4 = A(4)³ + B(4)² + C(4) + D = 0
a4 = A(64) + B(16) + C(4) + D = 0
a4 = 64A + 16B + 4C + D = 0
So we have the system of equations:
A + B + C + D = 10
8A + 4B + 2C + D = 8
27A + 9B + 3C + D = 3
64A + 16B + 4C + D = 0
Solve that system of equations and get
A = 5/6, B = -13/2, C = 35/3, D = 4
So
an = An³ + Bn² + Cn + D
becomes:
an = (5/6)n³ - (13/2)n² + (35/3)n + 4
Get a least common denominator of 6:
an = (5/6)n³ - (39/6)n² + (70/6)n + 24/6
Write the sum of the numerators over 6
an = (5*n³ - 39*n² + 70*n + 24)/6
Substitute 1 for n:
a1 = (5*1³ - 39*1² + 70*1 + 24)/6
a1 = (5*1 - 39*1 + 70*1 + 24)/6
a1 = (5 - 39 + 70 + 24)/6
a1 = 60/6
a1 = 10
Notice that the 1st term in your sequence was 10
Next substitute 2 for n:
a2 = (5*2³ - 39*2² + 70*2 + 24)/6
a2 = (5*8 - 39*4 + 70*2 + 24)/6
a2 = (40 - 156 + 140 + 24)/6
a2 = 48/6
a2 = 8
Notice that the 2nd term in your sequence was 8
Next substitute 3 for n:
a3 = (5*3³ - 39*3² + 70*3 + 24)/6
a3 = (5*27 - 39*9 + 70*3 + 24)/6
a3 = (135 - 351 + 210 + 24)/6
a3 = 18/6
a3 = 3
Notice that the 3rd term in your sequence was 3
Next substitute 4 for n:
a4 = (5*4³ - 39*4² + 70*4 + 24)/6
a4 = (5*64 - 39*16 + 70*4 + 24)/6
a4 = (320 - 624 + 280 + 24)/6
a4 = 0/6
a4 = 0
Notice that the 4th term in your sequence was 0
Next substitute 5 for n:
a5 = (5*5³ - 39*5² + 70*5 + 24)/6
a5 = (5*125 - 39*25 + 70*5 + 24)/6
a5 = (625 - 975 + 350 + 24)/6
a5 = 24/6
a5 = 4
Notice that the 5th term in your sequence was 4
Next substitute 6 for n:
a6 = (5*6³ - 39*6² + 70*6 + 24)/6
a6 = (5*216 - 39*36 + 70*6 + 24)/6
a6 = (1080 - 1404 + 420 + 24)/6
a6 = 120/6
a6 = 20
Notice that the 6th term in your sequence was 20
Next substitute 7 for n:
a7 = (5*7³ - 39*7² + 70*7 + 24)/6
a7 = (5*343 - 39*49 + 70*7 + 24)/6
a7 = (1715 - 1911 + 490 + 24)/6
a7 = 318/6
a7 = 53
Notice that the 7th term in your sequence was 53.
That was the last one given. Now we will continue on
to get some more terms
Next substitute 8 for n:
a8 = (5*8³ - 39*8² + 70*8 + 24)/6
a8 = (5*512 - 39*64 + 70*8 + 24)/6
a8 = (2560 - 2496 + 560 + 24)/6
a8 = 648/6
a8 = 108
The 8th term in your sequence is 108
Next substitute 9 for n:
a9 = (5*9³ - 39*9² + 70*9 + 24)/6
a9 = (5*729 - 39*81 + 70*9 + 24)/6
a9 = (3645 - 3159 + 630 + 24)/6
a9 = 1140/6
a9 = 190
The 9th term in your sequence is 190
Next substitute 10 for n:
a10 = (5*10³ - 39*10² + 70*10 + 24)/6
a10 = (5*1000 - 39*100 + 70*10 + 24)/6
a10 = (5000 - 3900 + 700 + 24)/6
a10 = 1824/6
a10 = 304
The 10th term in your sequence is 304
Next substitute 11 for n:
a11 = (5*11³ - 39*11² + 70*11 + 24)/6
a11 = (5*1331 - 39*121 + 70*11 + 24)/6
a11 = (6655 - 4719 + 770 + 24)/6
a11 = 2730/6
a11 = 455
The 11th term in your sequence is 455
Next substitute 12 for n:
a12 = (5*12³ - 39*12² + 70*12 + 24)/6
a12 = (5*1728 - 39*144 + 70*12 + 24)/6
a12 = (8640 - 5616 + 840 + 24)/6
a12 = 3888/6
a12 = 648
The 12th term in your sequence is 648
Next substitute 13 for n:
a13 = (5*13³ - 39*13² + 70*13 + 24)/6
a13 = (5*2197 - 39*169 + 70*13 + 24)/6
a13 = (10985 - 6591 + 910 + 24)/6
a13 = 5328/6
a13 = 888
The 13th term in your sequence is 888
Next substitute 14 for n:
a14 = (5*14³ - 39*14² + 70*14 + 24)/6
a14 = (5*2744 - 39*196 + 70*14 + 24)/6
a14 = (13720 - 7644 + 980 + 24)/6
a14 = 7080/6
a14 = 1180
The 14th term in your sequence is 1180
Next substitute 15 for n:
a15 = (5*15³ - 39*15² + 70*15 + 24)/6
a15 = (5*3375 - 39*225 + 70*15 + 24)/6
a15 = (16875 - 8775 + 1050 + 24)/6
a15 = 9174/6
a15 = 1529
The 15th term in your sequence is 1529
Edwin
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