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.

It's a recursive formula so start out calculating $f(2)$ using $f(1)$ and then move on to 3 and 4 — ClassicStyle, May 21 '14 at 23:15

Thomas · Accepted Answer · 2014-05-24T12:42:51.493

5

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

edited May 24 '14 at 12:42

answered May 21 '14 at 23:14

Thomas

43,555

@user152559: no problem. – Thomas May 21 '14 at 23:20
Ouch, it seems the downvote is mine. It must have been a missclick, and I haven't noticed it before. It's too old to remove, but if you could make some minor edit, I'd be happy to rectify this. Sorry for that! – Marcin Łoś May 24 '14 at 07:23
@MarcinŁoś: No problem. All forgiven. I edited. – Thomas May 24 '14 at 16:55
Great - thanks, changed to +1. – Marcin Łoś May 24 '14 at 17:09

score 3 · Answer 2 · answered May 21 '14 at 23:22

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 $$f(n)=16+(n+2)(n-1)$$

answered May 21 '14 at 23:22

score 0 · Answer 3 · answered May 21 '14 at 23:15

0

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

answered May 21 '14 at 23:15

Ross Millikan

374,822

score 0 · Answer 4 · answered May 22 '14 at 02:09

0

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

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

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

$f(1)=16$

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

answered May 22 '14 at 02:09

lhf

216,483

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

4 Answers4