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How does:

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

simplify into:

(1- (1/k))/2!

Could you show me step-by-step? thanks.

  • algebra -

    (k(k-1))/2!*(1/k)^2 square 1/k and group into one single faction

    = (k(k-1))/(2!*(k)^2) the k on top and the k on bottom cancel

    = (k-1)/(2!*k) you can rewrite this as:

    = (1/2!)*((k-1)/k) now multiply the whole thing by (1/k)/(1/k) {note: this is legal because it is the same as multiplying a number by 1}

    = (1/2!)*((1/k)(k-1))/((1/k)*k) the bottom cancel out {1/k*k = 1} and just multiply the top out

    = (1/2!)*(1-(1/k)) rewrite as:
    = (1-(1/k))/2!

    Fun problem!

  • algebra -

    What happened to the k in the 2!*k before you multiplied everything by (1/k)/(1/k)

  • algebra -

    (k(k-1))/2! * (1/k)^2
    = (k²-k)/k² / 2!
    = (k²/k² - k/k²) /2!
    = (1 - 1/k)/2!

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