How do I find $f(4)$ when $f(n)= f(n-1)+ 2n$?

Question

Can somebody please help me find $f(4)$ when $f(n)= f(n-1)+ 2n$? $f(1)$ equals $16$ by the way.

Answer 1

$$\begin{align} f(4) &= \color{blue}{f(3)} + 2\cdot 4 \\ &= \color{blue}{(f(2) + 2\cdot 3)} + 2\cdot 4 \\ &= \dots \end{align} $$

Answer 2

You could write $f(n)=g(n)+n^2-n$ if you are looking for a general solution. For $f(4)$, just compute

Answer 3

You can even find $f(n)$ for every $n$: in fact by telescoping $$f(n)-f(1)=\sum_{k=2}^n(f(k)-f(k-1))=\sum_{k=2}^n2k=(n+2)(n-1)$$ so $$f(n)=16+(n+2)(n-1)$$

Answer 4

$f(4)=f(3)+8$

$f(3)=f(2)+6$

$f(2)=f(1)+4$

$f(1)=16$

Sum all these and cancel to get $f(4)=8+6+4+16$.