# Sigma sequence problem

1. Feb 22, 2009

### UNknown 2010

1. The problem statement, all variables and given/known data
t1, t2, t3, t4 , ...... , tn is a sequence
t2 - t1 = t3 - t2 = t4 - t3 .............. = t(n-1) - tn

show that:

1/t1t2 + 1/t2t3 + 1/t3t4 ............... + 1/tn-1tn = n-1/t1tn

2. Relevant equations

3. The attempt at a solution
I tried to use sigma but I couldn't solve it ..

Please give me the beginning of the solution

2. Feb 22, 2009

### HallsofIvy

Staff Emeritus
Re: Sequenec..

Essentially this says that the sequence has a common difference and so is an "aritmetic" sequence. Let d= t2- t1= t3- t2= t4- t3 ... Then t2= t1+ d, t3= t2+ d= t1+ 2d, and, in general, tn= t1+ d(n-1).

1/t1t2= $$\frac{1}{t1(t1+ d)}= \frac{A}{t1}+ \frac{B}{t1+ d}$$
What are A and B?

1/t2t3= $$\frac{1}{(t1+d)(t1+2d)}= \frac{C}{t1+d}+ \frac{D}{t1+ 2d}$$
What are C and D?

Last edited: Feb 22, 2009
3. Feb 22, 2009

### UNknown 2010

Re: Sequenec..

:uhh:
what are a, b, c, d ???

4. Feb 22, 2009

### Unknot

Re: Sequenec..

Look up partial fraction.

5. Feb 22, 2009

### UNknown 2010

Re: Sequenec..

a = (-bt1+1)/(d+t)
b = [ -a (d+t1) + 1 ] / t1

6. Feb 22, 2009

### HallsofIvy

Staff Emeritus
Re: Sequenec..

??? To solve for a and b, there should be no "b" in the "a" formula or "a" in the "b" formula. Also don't use small letters for A and B. That's particularly confusing here since "d" and "D" represent different things.

$$\frac{1}{t1(t1+ d)}= \frac{A}{t1}+ \frac{B}{t1+ d}$$
multiply through by the denominator t1(t1+d):
1= A(t1+d)+ Bt1= (A+B)t1+ Ad
so we must have A+ B= 0 and Ad= 1. A= 1/d and B= -1/d

$$\frac{1}{t1t2}= \frac{1/c}\left(\frac{1}{t1}- \frac{1}{t1+d}\right)$$

For
$$\frac{1}{t2t3}=\frac{1}{(t1+d)(t1+2d)}= \frac{C}{t1+d}+ \frac{D}{t1+ 2d}$$
multiply through by (t1+d)(t1+2d) to get 1= C(t1+2d)+ D(t1+ d).

Now taking t1= -d, 1= dC, so C= 1/d and taking t1= -2d, 1= -dD, so D= -1/d.

[tex]\frac{1}{(t1+d)(t1+2d)}= \frac{1}{d}\left(\frac{1}{t1+d}- \frac{1}{t1+ 2d}\right)[/itex]

Putting those together,
[tex]\frac{1}{t1t2}+ \frac{1}{t2t3} = \frac{1}{c}\left(\frac{1}{t1}- \frac{1}{t1+d}+ \frac{1}{t1+d}- \frac{1}{t1+ 2d}\right)[/itex]

Do you see what happens? Can you do the same for the other terms?

Last edited: Feb 22, 2009