SOLUTION: Find the nth term for the following progression 8, 14, 22, 32, 44

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Question 1089382: Find the nth term for the following progression
8, 14, 22, 32, 44

Answer by MathLover1(20850) About Me  (Show Source):
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8, 14, 22, 32, 44
find differences:
8.........14.........22.........32.........44
.....6.............8.........10.........12
..............2.............2.........2-> second differences are constant, the sequence can be described by a quadratic formula of the form:
u%5Bn%5D=an%5E2%2Bbn%2Bc

now find coefficients a,+b, and c
To find the values of a, b and c consider the first 3 terms.
Using u%5B1%5D=8 and n=1 gives
8=a%2Bb%2Bc..........eq.1
Using u%5B2%5D=14 and n=2 gives
14=a%2A2%5E2%2Bb%2A2%2Bc
14=4a%2B2b%2Bc..........eq.2

Using u%5B3%5D=22 and n=3 gives
22=a%2A3%5E2%2Bb%2A3%2Bc
22=9a%2B3b%2Bc..........eq.3
solve the system:
8=a%2Bb%2Bc..........eq.1
14=4a%2B2b%2Bc..........eq.2
22=9a%2B3b%2Bc..........eq.3
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start with 8=a%2Bb%2Bc..........eq.1 and solve for c
c=8-b-a..........eq.1a
same with 14=4a%2B2b%2Bc..........eq.2
c=14-4a-2b .........eq.2b
from eq.1a and eq.2b we have
8-b-a=14-4a-2b
-b-a%2B4a%2B2b=14-8
3a%2Bb=6................solve for b
b=6-3a...........eq.3c
go to c=8-b-a..........eq.1a, substitute b

c=8-%286-3a%29-a
c=8-6%2B3a-a
c=+2%2B2a.........eq.4d
take eq3c and eq.4d, and substitute it in
22=9a%2B3b%2Bc..........eq.3
22=9a%2B3%286-3a%29%2B2%2B2a
22=9a%2B18-9a%2B2%2B2a
22=20%2B2a
22-20=2a
2=2a
a=1
find b=6-3a...........eq.3c
b=6-3%2A1
b=3
and c=+2%2B2a.........eq.4d will be
c=+2%2B2%2A1
c=+4
so, the nth term for the following progression is:
u%5Bn%5D+=+n%5E2+%2B+3n+%2B+4 (for all terms given)