Simplifying a Sigma Sequence Problem

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The discussion focuses on solving a problem involving an arithmetic sequence defined by the condition that the differences between consecutive terms are constant. The goal is to show that the sum of the reciprocals of the products of consecutive terms equals a specific fraction involving the first and last terms of the sequence. Participants explore the use of partial fractions to simplify the expressions, identifying constants A, B, C, and D in their equations. The solution process involves substituting values and manipulating the equations to reveal a telescoping nature in the series. The discussion emphasizes the importance of correctly defining variables and maintaining clarity in notation throughout the solution.
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Homework Statement


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

Homework Equations





The Attempt at a Solution


I tried to use sigma but I couldn't solve it ..




Please give me the beginning of the solution
 
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UNknown 2010 said:

Homework Statement


t1, t2, t3, t4 , ... , tn is a sequence
t2 - t1 = t3 - t2 = t4 - t3 ..... = t(n-1) - tn
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).

show that:

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

Homework Equations





The Attempt at a Solution


I tried to use sigma but I couldn't solve it ..




Please give me the beginning of the solution
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?
 
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:rolleyes:
what are a, b, c, d ?
 


Look up partial fraction.
 


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


? 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.

\frac{1}{(t1+d)(t1+2d)}= \frac{1}{d}\left(\frac{1}{t1+d}- \frac{1}{t1+ 2d}\right)[/itex]<br /> <br /> Putting those together,<br /> \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]&lt;br /&gt; &lt;br /&gt; Do you see what happens? Can you do the same for the other terms?
 
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