It is given that Xn = sum[ 1/k*(k+1)*(k+2)].

We can write 1/k*(k+1)*(k+2) as A / k + B/(k+1) + C/(k+2)

=> 1/k*(k+1)*(k+2) = A / k + B/(k+1) + C/(k+2)

=> 1 = A(k+1)(k+2) + B(k)(k+2) + C(k)(k+1)

=> 1 = A(k^2 + 3k + 2) + B(k^2 + 2k) + C(k^2 + k)

=> 1 = k^2( A + B + C) + k( 3A + 2B + C) + 2A

This gives A + B + C = 0, 3A + 2B + C = 0 and 2A = 1

=> A = 1/2

B+C + 1/2 = 0 and 3/2 + 2B + C = 0

=> 2B - B + C - C + 3/2 - 1/2 =0

=> B = -1

-1 + 1/2 + C =0

=> C = 1/2

So Xn = sum[ 1/2k - 1/(k+1) + 1/2(k+2)], where k = 1 to n.

=> Xn = (1/2)(1 + 1/2 + 1/3 + ... + 1/n) - (1/2 + 1/3... + 1/(n + 1)) + (1/2)(1/3 + ... + 1/n + 1/(n + 1))

=> Xn = (1/2)[1/2 - 1/(n + 1)(n + 2)]

**Therefore Xn = (1/2)[1/2 - 1/(n + 1)(n + 2)]**

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