How do you calculate this: 1/1*2*3 + 1/2*3*4 + ...+ 1/n(n+1)(n+2) ?

Answer 1

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

So if f(n) = 1/[n(n+1)] then you have each term in your sum being:

1/2[f(k)-f(k+1)]

So the sum is:

1/2 * (f(1)-f(2) + f(2)-f(3) + f(3)-f(4) ... + f(n)-f(n+1)) =
1/2 (f(1)-f(n+1))

f(1) = 1/2
f(n+1) = 1/[(n+1)(n+2)]

So the sum is:

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

=== added note:

m_s's solution is a nice general way to approach this sort of problem, but he fails to notice that you wanted the finite sum, not the limit. For the finite sum using his technique, you get:

A/1 + (B+A)/2 + (A+B+C)/3 ...
+ (A+B+C)/n + (B+C)/(n+1) + C/(n+2)

As with his argument, since A+B+C=0, only the outer terms survive:

A/1 + (B+A)/2 + (B+C)/(n+1) + C/(n+2)

A=1/2, B=-1, C= 1/2 gives us:

1/2 - 1/4 - 1/2(n+1) + 1/2(n+2)

which is equal to my answer:

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

Answer 2

1/n(n+1)(n+2) can be wriiten as A/n+B/(n+1)+C/(n+2)

so , RHS = A(n^2+3n+2)+B(n^2+2n)+C(n^2+n) / n(n+1)(n+2)

= n^2 (A+B+C) + n (3A+2B+C) + (2A) / n(n+1)(n+2)

so A = 1/2, B=-1, C=1/2

hence the given series can be written as:

(A/1+B/2+C/3) + (A/2+B/3+C/4) + (A/3+B/4+C/5) + ... (for infinite series)
=A+ 1/2(B+A) + 1/3(C+B+A)+1/4(C+B+A)+....
=A+1/2(B+A) since A+B+C=0
=1/2+1/2*(-1/2) = 1/2-1/4=1/4

take extra last 2 terms for finite series