# Thread: Help with these questions

1. ## Help with these questions

1.Prove that $n^4 + 4^n(n \geq2)$ is never a prime number.

2. Given $f(n) + f(n-1) = nf(n-1) + (n-1)f(n-2)(\mbox{for}\ n\ge 2)$
and f(0)=1 and f(1) =0, prove that
$\frac{f(n)}{n!}=\sum^n_{k=0}\frac{(-1)^k}{k!}$

3. In a given circle, C is the midpoint of arc ACD. CE is perpendicular to BD(B is a point on the circle between A and C).
If AB= $l_2$and BE = $l_1$, find ED in terms of $l_1$ and $l_2$.

2. For (1) you could try listing the first few values of the formula in decimal (this is always a good idea with this sort of problem). You might also like to try listing them in base 17.

3. #2 is not true

4. Originally Posted by malaygoel

2. Given $f(n) + f(n-1) = nf(n-1) + (n-1)f(n-2)(\mbox{for}\ n\ge 2)$
and f(0)=1 and f(1) =0, prove that
$\frac{f(n)}{n!}=\sum^n_{k=0}\frac{(-1)^k}{k!}$
The solution to the difference equation initial value problem:

$f(n) + f(n-1) = nf(n-1) + (n-1)f(n-2)(\mbox{for}\ n\ge 2)$

$f(0)=1$ and $f(1) =0$

is unique, so we need only show that $f(n)$ defined by:

$\frac{f(n)}{n!}=\sum^n_{k=0}\frac{(-1)^k}{k!}\ \ \ (1)$

is a solution.

------------------------------------------------------------

Now suppose:

$f(n)=n!\ \sum^n_{k=0}\frac{(-1)^k}{k!}$,

which is just a rewrite of $(1)$, and $n \ge 2$.

Then:

$
f(n)=n (n-1)!\ \left[\sum^{n-1}_{k=0}\frac{(-1)^k}{k!}+\frac{(-1)^n}{n!}\right]
$

$
=n\ f(n-1)+ \frac{n(n-1)!(-1)^n}{n!}=n\ f(n-1)+(-1)^n
$

Hence also:

$
f(n-1)=(n-1)\ f(n-2)+(-1)^{n-1}
$

So:

$
f(n)+f(n-1)=nf(n-1)+(n-1)f(n-2)
$
.

Also:

$
f(0)=0! (-1)^0/0!=1,\ \mbox{and}\ f(1)=1!((-1)^0/0!+(-1)^1/1!)=0
$
,

That is we have shown that $f(n)$ defined by $(1)$ satisfies the initial value problem
and hence is the unique solution.

RonL

5. Did you change the values of f(0) and f(1) around?

6. Originally Posted by ThePerfectHacker
Did you change the values of f(0) and f(1) around?
Not me, they were f(0)=1, f(1)=0 this morning when I first looked at the problem.

I did edit it this morning to make the TeX for "for x >= 2" clearer, that's
all honest Guv.

RonL