# Prove that the following formula is true for $n \geq 1$ by induction

Prove that the following formula is true for $n \geq 1$ by induction.

$a_{n} = a_{n-1} + 4n - 3 \\ a_{n} = 2n^{2} - n + 1 \\ a_{1} = 2$

My attempt follows below. I almost succeed in proving the formula for $n \geq 2$, but I fail (more specifically, I fail when I try to prove the formula for $n = k + 1$ and also, I do not know how to prove it for $n \geq 2$. Here is my attempt:

1. Let $n = 2 \\ 2 + 2*4 - 3 = 7 (a) \\ 2*2^{2} - 2 + 1 = 7 (b) \\ (a) = (b) \space QED$

2. Suppose the formula is true for $n = k$, which gives us: $\\ a_{k-1} + 4k - 3 = 2k^{2} - k +1 \\$ This leads to the formula being true for $n = k + 1$: $a_{k+1-1} + 4(k+1) -3 = 2(k+1)^{2} - (k+1) + 2$

$RHS = a_k + 4k -2 \\ LHS = 2(k^2 + 2k + 1) - k - 1 + 1 = 2k^2 +4k+2-k=(2k^2 - k +1) + 4k +1 =^{*} a_{k-1} + 4k -3 +4k+1 = a_{k-1} + 8k -2$

From the above you see that RHS is not equal to LHS... What am I doing wrong? And when it has been shown that RHS = LHS, how does one continue to show that the formula is true not only for $n \geq 2$ but also for $n \geq 1$?

*is where I use the assumption from (2)

For $n=1$

$2(1)^2-1+1=2=a_1$.

Assume this holds for $k$, that is $a_k=2k^2-k+1$

Now,

$a_{n+1}=a_{n}+4(n+1)-3$

$=2n^2-n+1+4(n+1)-3$

$=2n^2+4n+2 -n$

$=2(n+1)^2-n$

$=2(n+1)^2-(n+1)+1$

The argument follows by induction.

QED.

I find working with $n$ is better with $k$. So: $$a_n = a_{n-1} + 4n-3=2(n-1)^2-(n-1)+1+4n-3=2n^2-4n+2-n+2+4n-3=2n^2-n+1$$.