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# 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.

Please help me understand the Hence portion, I'm clueless as to what it's asking about. Your help would be very very much appreciated!! Thank you so much.

### 1 Answer

- DylanLv 51 decade agoFavorite Answer
(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