find an expression in terms n of the nth term of the following sequence 5,13,32,69,129,221......help
To find an expression in terms of n for the nth term of the given sequence, let's analyze the pattern and try to find a formula.
The sequence 5, 13, 32, 69, 129, 221 seems to be increasing in a non-linear manner.
Let's calculate the differences between consecutive terms:
- The difference between the 2nd and 1st terms is 13 - 5 = 8.
- The difference between the 3rd and 2nd terms is 32 - 13 = 19.
- The difference between the 4th and 3rd terms is 69 - 32 = 37.
- The difference between the 5th and 4th terms is 129 - 69 = 60.
- The difference between the 6th and 5th terms is 221 - 129 = 92.
From the differences, it appears that the differences themselves are increasing. This suggests that the sequence itself might be obtained by taking repeated differences.
Let's calculate the second differences:
- The difference between the 2nd difference and the 1st difference is 19 - 8 = 11.
- The difference between the 3rd difference and the 2nd difference is 37 - 19 = 18.
- The difference between the 4th difference and the 3rd difference is 60 - 37 = 23.
- The difference between the 5th difference and the 4th difference is 92 - 60 = 32.
The second differences are not constant, which indicates that the sequence might involve quadratic differences.
Let's calculate the third differences:
- The difference between the 2nd difference and the 1st difference is 11 - 8 = 3.
- The difference between the 3rd difference and the 2nd difference is 18 - 11 = 7.
- The difference between the 4th difference and the 3rd difference is 23 - 18 = 5.
The third differences are constant, suggesting that the nth term of the sequence might be a cubic polynomial of n.
To find this cubic polynomial, we can use the method of finite differences.
Let's start by defining the sequence's nth term as f(n). Since we have noticed that the third differences are constant, let's assume that the nth term can be expressed as a cubic polynomial.
f(n) = an^3 + bn^2 + cn + d
Now, let's substitute the first four terms into the equation:
For n = 1:
5 = a(1)^3 + b(1)^2 + c(1) + d
5 = a + b + c + d
For n = 2:
13 = a(2)^3 + b(2)^2 + c(2) + d
13 = 8a + 4b + 2c + d
For n = 3:
32 = a(3)^3 + b(3)^2 + c(3) + d
32 = 27a + 9b + 3c + d
For n = 4:
69 = a(4)^3 + b(4)^2 + c(4) + d
69 = 64a + 16b + 4c + d
We now have a system of four equations with four unknowns (a, b, c, d). We can solve this system using different methods such as substitution, elimination, or matrices. Once you solve for the values of a, b, c, and d, you will have the expression in terms of n for the nth term of the given sequence.