幫手計題m2數！plz！要步驟！20分！

a) 1/[(x+1)(x+2)] ≡ A/(x+1) + B/(x+2)1/[(x+1)(x+2)] ≡ [A(x+2) + B(x+1)] / [(x+1)(x+2)]1 = Ax + 2A + Bx + B1 = (A+B)x + 2A+BSoA+B = 0 ....(1)
2A+B = 1 ....(2)(2) - (1) :A = 1
B = - 1
b)n Σ r=1 1/[(r+1)(r+2)]=
n Σ r=1 1/(r+1) - 1/(r+2)=
(1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + ... + (1/n - 1/(n+1)) + (1/(n+1) - 1/(n+2))= 1/2 - 1/(n+2)


2011-01-16 13:03:55 補充：
=

n
---------
2(n+2)

a)
1/[(x+1)(x+2)]≡A/(x+1) +B/(x+2)
both sides times (x+1)(x+2),
A(x+2)+B(x+1) =1

sub x= -1,
A=1

sub x=-2,
-B =1
B = -1

Therefore A=1, B= -1

b)
n(色碼)r=1 1/[(r+1)(r+2)]
=n(色碼)r=1 [ 1/(r+1) - 1/(r+2) ]
=n+1(色碼)r=2 (1/r) - n+2(色碼)r=3(1/r)
=1/2 - 1/(n+2)
=n/[2(n+2)]

(a)
1/[(x+1)(x+2)]≡A/(x+1) +B/(x+2)
= [A(x+2) + B(x+1)] / (x+1)(x+2)

A(x+2) + B(x+1) = 1

When x = -1, A(-1+2) = 1
A = 1

When x = -2, B (-2 + 1) = 1
B (-1) = 1
B = -1

So, A = 1 and B = -1


(b)

From (a), 1/[(r+1)(r+2)] = 1/(x+1) - 1/(x+2)

(n) Σ (r=1) 1/[(r+1)(r+2)]
= (n) Σ (r=1) [1/(x+1) - 1/(x+2)]
= (n) Σ (r=1) 1/(x+1) - (n) Σ (r=1) 1/(x+2)
= (n) Σ (r=1) 1/(x+1) - (n+1) Σ (r=2) 1/(x+1)
= [1/(1+1) + (n) Σ (r=2) 1/(x+1) ] - {(n) Σ (r=2) 1/(x+1) + 1/[(n+1)+1)]}
= 1/(1+1) - 1/[(n+1)+1)]
= 1/2 - 1/(n+2)
= [(n+2) - 1] / 2(n+2)
= (n+1) / 2(n+2)