Evaluate the sum $\frac{3!+4!}{2(1!+2!)}+\frac{4!+5!}{3(2!+3!)}+\cdots+\frac{12!+13!}{11(10!+11!)}$
Problem
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Tags: factorial
rajai7
03.06.2014 19:12
(3!+4!)/{2(1!+2!)}={3!(1+4)}/{2!(1+2)} =5 From the series we can write:(3!+4!)/{2(1!+2!)}+(4!+5!)/{3(2!+3!)}+......+(12!+13!)/{11(11!+10!)} =5+6+.....+14 =95
sqing
04.06.2014 16:44
$\frac{(n+2)!+(n+3)!}{(n+1)(n!+(n+1)!)}=n+4.$
DIffCALCFTW
23.06.2016 04:06
We have $$\frac{(n+2)!+(n+3)!}{[(n+1)(n!+(n+1)!)]}=n+4.$$which means that the sum for this problem is $$(1+4)+(2+4)+(3+4)...+(10+4)$$which can be simplified to $$\frac{14*15}{2} - 10$$. Therefore our answer is $$105-10=95$$.
jeffisepic
23.06.2016 05:16
solution: use calculator
Flash12
23.06.2016 07:55
sqing wrote: $\frac{(n+2)!+(n+3)!}{(n+1)(n!+(n+1)!)}=n+4.$ Oops I don't get how that simplifies to $n+4$..
lucasxia01
23.06.2016 08:03
Just cancel the factorials. $\frac{(n+2)!+(n+3)!}{(n+1)(n!+(n+1)!)}=\frac{(n+1)!(n+2)(1 + n+3)}{(n+1)!(1 + n + 1)} = \frac{(n+2)(n+4)}{(n+2)} = n+4$.