Given a/(k+1)!  -b/k! +c/(k-1)!=(k^2-k-1)/(k+1)!, what are natural numbers a,b,c?  

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll write the denominators:

 (k+1)! = (k-1)!*k*(k+1)

k! = (k-1)!*k

We'll re-write the identity from enunciation:

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

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

a - b(k+1) + c*k(k+1) = (k^2-k-1)

We'll remove the brackets:

a - bk  - b + ck^2 + ck = k^2 - k - 1

We'll combine like terms from the left side:

ck^2 + k(c - b) + a - b = k^2 - k - 1

Comparing, we'll get:

c = 1

c - b = -1

b = c + 1 => b  = 2

a - b = -1

a = b - 1

a = 1 

The natural numbers a,b,c are: a = 1, b  = 2 and c = 1, so that:

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

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