Question:
The recursive function [tex]f(0) = 1[/tex], [tex]f(n) = f(n-1) + 2n[/tex] represents the nth term of a sequence. Determine the explicit function
Answer:
[tex]f(n) = n^2 + n+1[/tex]
Step-by-step explanation:
Given
[tex]f(0) = 1[/tex]
[tex]f(n) = f(n-1) + 2n[/tex]
Required
Write an explicit formula
Let n = 1
[tex]f(1) = f(1-1) + 2*1[/tex]
[tex]f(1) = f(0) + 2[/tex]
[tex]f(1) = 1 + 2 = 3[/tex]
Let n = 2
[tex]f(2) = f(2-1) + 2*2[/tex]
[tex]f(2) = f(1) + 4[/tex]
[tex]f(2) = 3 + 4 = 7[/tex]
Let n =3
[tex]f(3)=f(3-1) + 2 * 3[/tex]
[tex]f(3)=f(2) + 6[/tex]
[tex]f(3)=7+ 6 = 13[/tex]
Let n = 4
[tex]f(4) = f(4 - 1) + 2 * 4[/tex]
[tex]f(4) = f(3) + 8[/tex]
[tex]f(4) = 13+ 8 = 21[/tex]
So, we have:
[tex]f(1) = 3= 1 + 2 * 1=1+0*1*2*1[/tex]
[tex]f(2) = 7= 3 + 2 * 2=1+1*2+2*2[/tex]
[tex]f(3) = 13= 7 + 2 * 3= 1+2*3+2*3[/tex]
[tex]f(4) = 21 = 13 + 2 * 4= 1+3*4 +2 *4[/tex]
Following the above pattern:
[tex]f(n) = 1 + (n - 1) * n + 2 * n[/tex]
[tex]f(n) = 1 + n(n - 1) + 2n[/tex]
Open bracket
[tex]f(n) = 1 + n^2 - n + 2n[/tex]
[tex]f(n) = 1 + n^2 + n[/tex]
[tex]f(n) = n^2 + n+1[/tex]
