# How do you prove the following?

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

Dec 8, 2017

Kindly refer to a Proof given in the Explanation.

#### Explanation:

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

$= \left\{1.2 .3 \ldots . \left(n - 1\right)\right\} n ,$

 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!

Dec 8, 2017

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

#### Explanation:

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!)