Answer:
Proved
Step-by-step explanation:
Given
[tex]1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)[/tex]
[tex]n \geq 1[/tex]
Required
Prove by induction
[tex]1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)[/tex]
Increment n by 1 on both sides
[tex]1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+1+1)(2(n+1)+1)[/tex]
Simplify the right hand side
[tex]1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+2)(2n+2+1)[/tex]
[tex]1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+2)(2n+3)[/tex]
Group the left hand side
[tex](1+5+9+...+[4(n-1)+1]+[4n+1])+[4(n+1)+1] = (n+2)(2n+3)[/tex]
Recall that
[tex]1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)[/tex] ----[Given]
So; Substitute [tex](n+1)(2n+1)[/tex] for [tex]1+5+9+...+[4(n-1)+1]+[4n+1][/tex] on the left hand side
[tex](n+1)(2n+1)+[4(n+1)+1] = (n+2)(2n+3)[/tex]
Open All Brackets
[tex]2n^2 + n + 2n + 1 + 4n + 4 + 1 = 2n^2 + 3n + 4n + 6[/tex]
Collect Like Terms
[tex]2n^2 + n + 2n + 4n+ 1 + 4 + 1 = 2n^2 + 3n + 4n + 6[/tex]
[tex]2n^2 + 7n+ 6 = 2n^2 + 7n + 6[/tex]
Notice that the expression on both sides are equal;
Hence, the given expression has been proven
