Figure-It-Out Function! - Part 3: The Powers of Factorials

Algebra Level 4

Let there be a function $f(x):\mathbb{Z}\rightarrow \mathbb{Z}$ with the following conditions

$\begin{cases}{f(x+1)=f(x)+(f(x))!} \\ {f(3)=28}\end{cases}$

Evaluate $\displaystyle \sum_{i=-1}^{4} \left [ f(i) \cdot f(3-i) \right ]$

1 solution

Shivamani Patil
Aug 17, 2014

Now let $x=2$

Therefore $f(3)=f(2)+(f(2))!$

Now it is interesting to note that $f(3)=28=4+4!$

Therefore $4+4!=f(2)+(f(2))!$

By comparing

we get $f(2)=4$ ...... $1$

Now let $x=1$

Therefore we get $f(2)=f(1)+(f(1))!$

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

Again by comparing we get $f(1)=2$

By continuing above process we get $f(0)=1$ ..... $2$

By condition we get $f(-1)=0$ ...... $3$

That summation is equal to $2[f(-1)f(4)+f(1)f(2)+f(0)f(3)]$

By condition $f(4)=28+28!$

But we don't need to evaluate it

Therefore summation equals $2[0+8+28]$

i.e $72$

great problem man.................keep it up............

Gaurav Kashyap - 6 years, 9 months ago

Very good problem which requires logical thinking.

shivamani patil - 6 years, 10 months ago