# Given the string xn=Sum 1/k(k+1)(k+2) where k = 1 to n , what is xn ?

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### 3 Answers

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)]**

To determie the general terms of the string xn, we'll have to decompose the given fraction into elementary ratios first.

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

We'll multiply both sides by k(k+1)(k+2):

1 = A(k+1)(k+2) + Bk(k+2) + Ck(k+1)

We'll remove the brackets:

1 = Ak^2 + 3Ak + 2A + Bk^2 + 2Bk + Ck^2 + Ck

We'll combine and factorize like terms from the right side:

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

Comparing, we'll get:

2A =1

**A = 1/2**

A+B+C = 0

B + C = -1/2 (1)

3A + 2B + C = 0

2B + C = -3/2 (2)

We'll subtract (1) from (2):

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

We'll combine and eliminate like terms:

**B = -1**

C = -1/2 - B

C = 1 - 1/2

**C = 1/2**

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

Now, we'll determine the sum:

For k = 1:

1/2 - 1/2 + 1/6

For k = 2:

1/4 - 1/3 + 1/8

For k = 3:

1/6 - 1/4 + 1/10

For k = 4:

1/8 - 1/5 + 1/12

........................

For k = n:

1/2n - 1/(n + 1) + 1/2(n + 2)

We'll add all terms and we'll get:

(1/2)(1 + 1/2 + 1/3 + ... + 1/n) - 1 - 1/2 - (1/3 + ... + 1/n + 1/(n + 1)) - + (1/2)(1/3 + ... + 1/n + 1/(n + 1))

Combining the terms, we'll get:

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

xn = Sum {1/k(k+1)(k+2)}, k = 1,2, 3,.... n.

Therefore xn = 1/1*2*3 +1/2*3*4+1*3*4+5+....+1/n(n+1)(n+2).

Consider the term 1/k(k+1)-1/(k+1)(k+2) = (k+2 -k)/k(k+1)(k+2) = 2/k(k+1)(k+2).

Therefore 1/k(k+2)(k+2) = (1/2){1/k(k+1) - 1/(k+1)(k+2)}.

So now we can rewrite the given sum xn as below splitting each term on the right side.:

xn = {1/2){(1/1*2 - 1/2*3) +(1/2*3-1/3*4)+1/3*4-1/4*5)+...1/n(n+1)-1/(n+1)(n+2)}.

xn = 1/2) { 1/1*2 -1/2*3+1/2*3 -1/3*4+1/3*4-1/4*5+1/4*5-.......-1/n(n+1)+1/n(n+1)-1/(n+1)(n+2)}.

xn = (1/2){ 1/2 - 1/(n+1)(n+1)}.

xn = (1/2){ 1/2 - 1/(n+1)(n+2)}.