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$f(n)=f(n-1)+2n-1$
$f(n)=f(n-1)+2(n-1)$
Let us just focus on the right hand side.
$f(n-1)+2(n-1)$
$=f(n-2)+2(n-2)+2(n-1)$
$=f(n-2)+2((n-2)+(n-1)$
$=f(n-3)+2((n-3)+(n-2)+(n-1))$
This process goes on,
$=f(n-k)+2((n-k)+(n-k+1)+\cdots +(n-2)+(n-1))$
Now to find the value of $f(n)$ , we can only use $f(1)=2$. So to get f(1) on the right hand side we need to get $k=n-1$
$=f(n-n+1)+2((n-n+1)+(n-n+2)+\cdots +(n-2)+(n-1))$
$\therefore f(n)=f(1)+2(1+2+3+ \cdots +n-2+n-1)$[Note that this makes our problem very easy]
So,
$f(100)=f(1)+2(1+2+3+\cdots +98+99)$
$f(100)=2+2(\frac{99\times 98}{2})$
$f(100)=2+2(4851)$
$\therefore f(100)=9704$

$f(100)=f(1)+2(1+2+3+\cdots +98+99)$
$f(100)=2+2(\frac{99\times 100}{2})$
$f(100)=2+2(4950)$
$\therefore f(100)=9902$