What is the sum of 1/1*2*3 + 1/2*3*4 + 1/3*4*5 + .... + 1/n*(n+1)*(n+2)?

Question

What is the sum of 1/1*2*3 + 1/2*3*4 + 1/3*4*5 + .... + 1/n*(n+1)*(n+2)?

Answer 1

Let Tr = 1/[r*(r+1)*(r+2)]



Using partial fractions:



Let Tr = [A/r]+[B/(r+1)]+[C/(r+2)] = [A(r+1)(r+2)+B(r)(r+2)+C(r)(r+1)]/[r(r+1)(r+2)]



Then 1 = [A(r+1)(r+2)+B(r)(r+2)+C(r)(r+1)]



Substituting r as 0, -1, and -2, A, B and C can be found.



A=1/2

B= -1

C= 1/2



Use this to express the first three and the last three terms of the progression.



I've spaced it out so that you can see the pattern:



T1 = 1/2 - 1/2 + 1/(2*3)

T2 = 1/(2*2) - 1/3 + 1/(2*4)

T3 = 1/(2*3) - 1/4 + 1/(2*5)

T4 = 1/(2*4) - 1/5 + 1/(2*6)

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

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

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

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





On adding all the terms, all the terms except for six terms cancel out.

The upper left triangle and the lower right triangle.



You get:



S= (1/2) - (1/2) + (1/4) + 1/[2*(n+1)] - 1/(n+1) + 1/[2*(n+2)]

S= 1/4 - 1/[2*(n+1)] +1/[2*(n+2)]



Which equates to:

S= 1/4 - 1/[(n+1)(n+2)]

That's the sum for a given value of n.

If n tends to infinity, S becomes equal to 1/4=0.25

Answer 2

Let Tr = 1/[r*(r+1)*(r+2)]



Using partial fractions:



Let Tr = [A/r]+[B/(r+1)]+[C/(r+2)]

we know another equation

(n+1)^2-n(n+2)=1

then we have this equation:(n+1)/n(n+2)-1/(n+1)=1/n(n+1)(n+2)

with the equation: Tr= [A/r]+[B/(r+1)]+[C/(r+2)]

you get B=-1, then we have new equation:

(n+1)/n(n+2)= [A/r]+[C/(r+2)]

[A/r]+[C/(r+2)] =A(r+2)+C(r)/r(r+2)=(n+1)/n(n+2)

we substituting n as r, you get: A(r+2)+C(r)=r+1, A(r+2)+C(r)=[r+(r+2)]/2

we found: A=C=1/2, B=-1



Spaced it out so that you can see the pattern:



T1 = 1/2 - 1/2 + 1/(2*3)

T2 = 1/(2*2) - 1/3 + 1/(2*4)

T3 = 1/(2*3) - 1/4 + 1/(2*5)

T4 = 1/(2*4) - 1/5 + 1/(2*6)

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

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

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

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





On adding all the terms, all the terms except for six terms cancel out.

The upper left triangle and the lower right triangle.



You get:



S= (1/2) - (1/2) + (1/4) + 1/[2*(n+1)] - 1/(n+1) + 1/[2*(n+2)]

S= 1/4 - 1/[2*(n+1)] +1/[2*(n+2)]

Answer 3

S_1 = 1 / 6 S_2 = 1 / 6 + 1 / 24 = 5 / 24 S_3 = 1 / 6 + 1 / 24 + 1 / 60 = ? Can you make a formula for the sum based on the partial sums??