If f(x)=f(x-1)+5x and f(1)=2 then how to find f(20)?

[AlphaCentauri]

If f(x)=f(x-1)+5x and f(1)=2 then how to find f(20)?

 

[Deleted member 4993]

If f(x)=f(x-1)+5x and f(1)=2 then how to find f(20)?

Easy but long way:

f(2) = f(1) + 5 * 2 = 12 ........ [edited .... Thanks Jeff]

f(3) = f(2) + 5 * 3 = ?

and continue.....

There are other methods - what method have you been taught?

 

[MarkFL]

I would treat it as a linear inhomogeneous difference equation to get the closed form for $\displaystyle f(x)$.

 

[MarkFL]

Another method would be to observe:

$\displaystyle \displaystyle f(x)=f(1)+5\sum_{k=2}^{x}(k)=5\sum_{k=1}^{x}(k)+(f(1)-5)=\,?$

 

[TomHE]

I would treat it as a linear inhomogeneous difference equation to get the closed form for $\displaystyle f(x)$.

Expanding on MarkFL's suggestion, $\displaystyle f(x) = f(x-1) + 5x, \; f(1)=2$ yields the recursively defined sequence:
$\displaystyle a_n = a_{n-1} + 5n$ whose first term is $\displaystyle a_1 = 2$. Note that $\displaystyle f(n) = a_n$ when $\displaystyle n$ is a natural number.
With the goal of finding a closed form formula for $\displaystyle a_n$ in mind, the first few terms are:
$\displaystyle a_1 = 2$
$\displaystyle a_2 = a_1 + 5 \cdot 2$
$\displaystyle a_3 = a_2 + 5 \cdot 3 = (a_1 + 5 \cdot 2) + 5 \cdot 3 = a_1 + 5(2 +3)$
$\displaystyle a_4 = a_3 + 5 \cdot 4 = (a_1 + 5(2 +3)) + 5 \cdot 4 = a_1 + 5(2+3+4)$
By inspection, a formula (whose validity can be proved by an induction argument) for $\displaystyle a_n$ in terms of $\displaystyle a_1$ and $\displaystyle n$ is:
$\displaystyle a_n = a_1 + 5(2 + 3 + \cdots + n)$
A closed form formula for $\displaystyle a_n$ can be derived from the identity $\displaystyle 1+2+\cdots+n=\frac{n(n+1)}2$ and the fact that $\displaystyle a_1 = 2$.