SOLUTION

The sequence is not linear nor quadratic , hence it is cubic sequence. We knew that it is a cubic sequence when we subtract the terms and the third difference is a constant.

The nth terms of the cubic sequence will be the cubic polynomial:

» an³+bn²+cn+d

We will formulate systems of equations:

If n = 1

» a(1)³+b(1)²+c(1)+d = 1

» a + b + c + d = 1 -eq1

If n = 2

» a(2)³+b(2)²+c(2)+d = 8

» 8a + 4b + 2c + d = 8 -eq2

If n = 3

» a(3)³+b(3)²+c(3)+d = 27

» 27a + 9b + 3c + d = 27 - eq3

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We just formulated 3 different equations and we will solve each variables to get the explicit formula:

eq2 - eq1

8a+4b+2c+d = 8

- a + b + c + d = 1

= 7a + 3b + c = 7 - eq1

eq3 - eq2

27a + 9b + 3c = 27

- 8a + 4b + 2c = 8

= 21a + 5b + c = 21 - eq2

eq3 - eq1

27a + 9b + 3c + d = 27

- a + b + c + d = 1

= 26a + 8b + 2c = 26 - eq3

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Solving by elimination: Eliminating a

-3( 7a + 3b + c = 7 )

21a + 5b + c = 21

It follows that

-21a - 9b - 3c = -21

+ 21a - 5b + c = 21

= -14b - 2c = 0

Then finding the value of b and c

Notice , that if -14b -2c = 0 , then probably b and c are also zero because 0 - 0 = 0

a = ? , b = 0 , c = 0

So , we will substitute the values of b and c to get the value of a.

» Finding the value of a

21a - 5b + c = 21

21a - 5(0) + 0 = 21

21a = 21

a = 21/21

a = 1

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Finally we will arrive with an explicit formula:

a = 1 , b = 0 , c = 0 , d = 0

an = an³+bn²+cn+d

an = 1(n³) + 0(n²) + 0(n) + 0

an = n³ + 0 + 0 + 0

an = n³

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Checking: an = n³

For n = 1

• an = n³
• a1 = 1³
• a1 = 1

For n = 2

• an = n³
• a2 = 2³
• a2 = 8

For n = 3

• an = n³
• a3 = 3³
• a3 = 9

For n = 4

• an = n³
• a4 = 4³
• a4 = 64

For n = 5

• an = n³
• a5 = 5³
• a5 = 125

Therefore , our formula is correct:

ANSWER

• an = n³

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