partial fractions

ryan750

2

express (3r - 2)/[r(r+1)(r+2)] in partial fractions

pls help i keep getting stuck - thanx

Reply 1

//Mike

Split it up into three different fractions with denominators r, r+1 and r+2. Since all of the denominators are of first order the numerators will be A, B and C respectively. You got that far?

Reply 2

ryan750

OP

2

//Mike

Split it up into three different fractions with denominators r, r+1 and r+2. Since all of the denominators are of first order the numerators will be A, B and C respectively. You got that far?

yup

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i then get 3 equations:

-2 = b + c/2
-5 = -a + c
-8 = -a/2 - b

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is that right?

Reply 3

//Mike

(3r - 2)/[r(r+1)(r+2)] ≡ [A/r] + [B/(r+1)] + [C/(r+2)]

Therefore 3r - 2 = A(r+1)(r+2) + B(r)(r+2) + C(r)(r+1)
3r-2 = A(r²+3r+2) + B(r²+2r) + C(r²+r)

Then compare coefficients:
r²: 0 = A + B + C (eqn1)
r: 3 = 3A + 2B + C (eqn2)
constant: -2 = 2A (eqn3)

Solve the three simultaneously: From eqn 3 we know A = -1; therefore eqn1 becomes 0= B+C-1 (eqn4) and eqn2 becomes 3=2B+C-3 (eqn5)
From eqn4; B+C=1 and eqn5; 2B+C=6. Substituting B = 1-C into eqn5 gives 2-2C+C = 6; 2-C=6; C=-4, therefore B = 1--4 = 5

Then substitute the values A,B,C into the original identity.

Reply 4

ryan750

OP

2

thanx so much i get it now

No problem :smile:

OP

the next part says - hence find the sum of the first n terms of the series

1/1x2x3 + 4/2x3x4 + 7/3x4x5 + ...

im rubbish at series - ne ideas?

Reply 7

ryan750

OP

2

i found how to do it - its to do with cancelling of the series and only the first few and last few count. or sumthing sumthing

Reply 8

//Mike

Well the general formula for the series is (3r - 2)/[r(r+1)(r+2)]; and we've just expressed this as [5/(r+1)] - [1/r] - [4/(r+2)] in partial fractions. Can't remember what the formula for summation to n is, but I think you bung it all in and get an answer...

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That is probably the most unuseful bit of advice I think I could have given...

partial fractions

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