given only the first 4 terms, I notice that the common difference forms an
arithmetic sequence, so the nth term is a quadratic.
let's look for a pattern: double each term , then write in factor form
2 --- 4 ---- (1+3)(1-2) = -4
0 --- 0 --- (2+3)(2-2) = 0
-3 --- -6 --- (3+3)(3-2) = 6
-7 --- -14 -- (4+3)(4-2) = 14 <----- notice the results are twice the opposites
of our starting values.
so reversing the steps, I would say
term(n) = (-1/2)(n + 3)(n - 2)
What is nth term of 2, 0, -3, -7?
2 answers
mathhelper's solution is correct.
You can calculate it like this.
Interpolation method.
With the interpolation polynomial, the degree of the polynomial is one less than the number of points.
When you have four points, the interpolation polynomial is of the third degree.
Interpolation polynomial is:
tn = a n³ + b n² + c n + d
Put n = 1 , n = 2 , n = 3 and n = 4 in this interpolation polynomial.
n = 1 , t1 = 2
a ∙ 1³ + b ∙ 1² + c ∙ 1 + d = 2
a + b + c + d = 2
n = 2 , t2 = 0
a ∙ 2³ + b ∙ 2² + c ∙ 2 + d = 0
8 a + 4 b + 2 c + d = 0
n = 3 , t3 = - 3
a ∙ 3³ + b ∙ 3² + c ∙ 3 + d = - 3
27 a + 9 b + 3 c + d = - 3
n = 4 , t4 = - 7
a ∙ 4³ + b ∙ 4² + c ∙ 4 + d = - 7
64 a + 16 b + 4 c + d = - 7
Now solve system of four equation.
a + b + c + d = 2
8 a + 4 b + 2 c + d = 0
27 a + 9 b + 3 c + d = - 3
64 a + 16 b + 4 c + d = - 7
The solution is:
a = 0 , b = - 1 / 2 , c = - 1 / 2 , d = 3
Interpolation polynomial is:
tn = 0 ∙ n³ - 1 / 2 n² - 1 / 2 n + 3
tn = - 1 / 2 n² - 1 / 2 n + 3
tn = - 1 / 2 ( n² + n - 6 )
Factorization gives:
tn = - 1 / 2 ( n + 3 ) ( n - 2 )
Calculate the value of this polynomial for n = 5, n = 6, n = 7...
n = 5 , t5 = - 12
n = 6 , t6 = -18
n = 7 , t7 = - 25
etc.
You can calculate it like this.
Interpolation method.
With the interpolation polynomial, the degree of the polynomial is one less than the number of points.
When you have four points, the interpolation polynomial is of the third degree.
Interpolation polynomial is:
tn = a n³ + b n² + c n + d
Put n = 1 , n = 2 , n = 3 and n = 4 in this interpolation polynomial.
n = 1 , t1 = 2
a ∙ 1³ + b ∙ 1² + c ∙ 1 + d = 2
a + b + c + d = 2
n = 2 , t2 = 0
a ∙ 2³ + b ∙ 2² + c ∙ 2 + d = 0
8 a + 4 b + 2 c + d = 0
n = 3 , t3 = - 3
a ∙ 3³ + b ∙ 3² + c ∙ 3 + d = - 3
27 a + 9 b + 3 c + d = - 3
n = 4 , t4 = - 7
a ∙ 4³ + b ∙ 4² + c ∙ 4 + d = - 7
64 a + 16 b + 4 c + d = - 7
Now solve system of four equation.
a + b + c + d = 2
8 a + 4 b + 2 c + d = 0
27 a + 9 b + 3 c + d = - 3
64 a + 16 b + 4 c + d = - 7
The solution is:
a = 0 , b = - 1 / 2 , c = - 1 / 2 , d = 3
Interpolation polynomial is:
tn = 0 ∙ n³ - 1 / 2 n² - 1 / 2 n + 3
tn = - 1 / 2 n² - 1 / 2 n + 3
tn = - 1 / 2 ( n² + n - 6 )
Factorization gives:
tn = - 1 / 2 ( n + 3 ) ( n - 2 )
Calculate the value of this polynomial for n = 5, n = 6, n = 7...
n = 5 , t5 = - 12
n = 6 , t6 = -18
n = 7 , t7 = - 25
etc.
1 answer