Question

Let n>=1. Prove by induction for all n>=1 1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1).

Answer 1

Answer:

Proved

Step-by-step explanation:

Given

$1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)$

$n \geq 1$

Required

Prove by induction

$1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)$

Increment n by 1 on both sides

$1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+1+1)(2(n+1)+1)$

Simplify the right hand side

$1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+2)(2n+2+1)$

$1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+2)(2n+3)$

Group the left hand side

$(1+5+9+...+[4(n-1)+1]+[4n+1])+[4(n+1)+1] = (n+2)(2n+3)$

Recall that

$1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)$ ----[Given]

So; Substitute $(n+1)(2n+1)$ for $1+5+9+...+[4(n-1)+1]+[4n+1]$ on the left hand side

$(n+1)(2n+1)+[4(n+1)+1] = (n+2)(2n+3)$

Open All Brackets

$2n^2 + n + 2n + 1 + 4n + 4 + 1 = 2n^2 + 3n + 4n + 6$

Collect Like Terms

$2n^2 + n + 2n + 4n+ 1 + 4 + 1 = 2n^2 + 3n + 4n + 6$

$2n^2 + 7n+ 6 = 2n^2 + 7n + 6$

Notice that the expression on both sides are equal;

Hence, the given expression has been proven