# Express (7r+10)/r(r+1)(r+2) in partial fractions?

Hence find Summation (7r+10)/r(r+1)(r+2) where first term has r=1 and last term is n, giving your answer in the form M-f(n) where M is a constant.

Deduce the exact value of Summation (21r+30)/r(r+1)(r+2), where first term has r=22 and last term is infinity.

Relevance
• Dylan
Lv 5

(7r + 10)/r(r + 1)(r + 2) = A/r + B/(r + 1) + C/(r + 2)

A(r + 1)(r + 2) + Br(r + 2) + Cr(r + 1) = 7r + 10

r = 0 --> 2A = 10 --> A = 5

r = -1 --> -B = 3 --> B = -3

r = -2 --> 2C = -4 --> C = -2

Complete fraction:

(7r + 10)/r(r + 1)(r + 2) = 5/r - 3/(r + 1) - 2/(r + 2)

S(n) = 17/1*2*3 + 24/2*3*4 + 31/3*4*5 + ...

+ [7(n - 1) + 10]/(n - 1)n(n + 1) + (7n + 10)/n(n + 1)(n + 2)

= [5/1 - 3/2 - 2/3] + [5/2 - 3/3 - 2/4] + [5/3 - 3/4 - 2/5] + ...

+ [5/(n - 1) - 3/n - 2/(n + 1)] + [5/n - 3/(n + 1) - 2/(n + 2)]

Notice that the terms with the same denominators cancel

= 5 + 2/2 - 5/(n + 1) - 2/(n + 2)

= 6 - (7n + 12)/(n + 1)(n + 2)

Do the same for the last part. The last term is r = n and n approaches infinity

(21r + 30)/r(r + 1)(r + 2) = 15/r - 9/(r + 1) - 6/(r + 2)

S(n) = (21*22 + 30)/22*23*24 + (21*23 + 30)/23*24*25 + (21*24 + 30)/24*25*26 + ...

+ [(21(n - 1) + 30]/(n - 1)n(n + 1) + (21n + 30)/n(n + 1)(n + 2)

= [15/22 - 9/23 - 6/24] + [15/23 - 9/24 - 6/25] + [15/24 - 9/25 - 6/26] + ...

+ [15/(n - 1) - 9/n - 6/(n + 1)] [15/n - 9/(n + 1) - 6/(n + 2)

= 15/22 + 6/23 - 15/(n + 1) - 6/(n + 2)

= 477/506 - 15/(n + 1) - 6/(n + 2)

lim S(n) (n --> ∞)

= 477/506 - 15/∞ - 6/∞

= 477/506 - 0 - 0

= 477/506