# Factorial applicationGiven X/(k+1)! -Y/k! +Z/(k-1)!=(k^2-k-1)/(k+1)!, what are natural numbers X,Y,Z?

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Student Comments

giorgiana1976 | Student

We'll write the denominators:

(k+1)! = (k-1)!*k*(k+1)

k! = (k-1)!*k

We'll re-write the identity from enunciation:

x/(k-1)!*k*(k+1) - y/(k-1)!*k + z/(k-1)! = (k^2-k-1)/(k-1)!*k*(k+1)

We'll multiply both sides by (k-1)!*k*(k+1):

x - y(k+1) + z*k(k+1) = (k^2-k-1)

We'll remove the brackets:

x - yk - y + zk^2 + zk = k^2 - k - 1

We'll combine like terms from the left side:

zk^2 + k(z - y) + x - y = k^2 - k - 1

Comparing, we'll get:

**z = 1**

z - y = -1

y = z + 1 => **y = 2**

x - y = -1

x = y - 1

**x = 1 **

**The natural numbers x,y,z are: x = 1, y = 2 and z = 1, so that:**

**1/(k+1)! - 2/k! + 1/(k-1)! = (k^2-k-1)/(k+1)!**