Find the value of n that satisfies n-(n+1)!+6n!=n!+4, where n!=n*(n-1)*(n-2)***2*1 .
n ==1 and
n == 4
Both of them satisfy the equation.
I'd be interested in seeing a formal solution for this.
CPhill presents a clear solution here:
https://web2.0calc.com/questions/pls-help-asap_40
GA
Thanks Ginger, but the questions are not the same.
\(n-(n+1)!+6n!=n!+4\\ n-n!(n+1)+5n!=4\\ n+n!(-n-1)+5n!=4\\ n+n!(-n-1+5)=4\\ n!(4-n)=4-n\\ n!(4-n)-(4-n)=0\\ (n!-1)(4-n)=0\\ n!=1\qquad or \qquad n=4\\ n=0,\;\;1\;\;or\;\;4\)
Thank you for the solution.
I tried this problem yesterday and could not get it.
n - (n+1)n! + 5n! = 4
n + (4 - n)n! = 4
I did not think of factoring it from here, that's a really nice solution.
=^._.^=
Nice solution , Melody !!!!!
Thanks guys
