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]

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