How do you prove the following?

Question

#1/((n-1)!)# + #1/((n-2)!)# = #n^2/(n!)#

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Answer 1 · 2017-12-08T04:08:31

Kindly refer to a Proof given in the Explanation.

Recall that, #n! =1.2.3.....(n-2)(n-1)n,#

#={1.2.3....(n-1)}n,#

# rArr n! =n.(n-1)!..................................................<<1>>.#

Similarly, #n! = n(n-1).(n-2)!.............................<<2>>.#

Using this, we have,

#"The Expression="1/((n!)/n) + 1/{(n!)/(n(n-1))},#

#=n/(n!)+{n(n-1)}/(n!),#

#={n+n(n-1)}/(n!),#

#={n+n^2-n}/(n!),#

#n^2/(n!).#

Q.E.D.

Enjoy Maths!

Answer 2 · 2017-12-08T04:20:47

To prove the equality we need to add up the two fractions

To add the two fractions on the left hand side we should first observe that:

#(n-1)! = (n-2)! * (n-1) #

So we can write

#1/((n-1)!)+1/((n-2)!) = 1/((n-2)! * (n-1)) + 1/((n-2)!) = (1+(n-1))/((n-2)! * (n-1)) = n/((n-2)! * (n-1))#

Now if we multiply both top and bottom by #n#, we get:

#(n*n)/((n-2)! * (n-1) * n) = n^2 /((n-2)! * (n-1) * n)#

but the expression at the bottom

#(n-2)! * (n-1) * n= n!#, so we end up with the equality

#1/((n-1)!)+1/((n-2)!) = n^2 / (n!) #