How Can You Solve the Problem of P1Q1 + P2Q2 + ... in Binomial Expansions?

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving the summation of products of sequences defined as P_r and Q_r, specifically in the context of binomial expansions. The original poster seeks a method to simplify or evaluate the expression P1Q1 + P2Q2 + ... + Pn-1Qn-1.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to express the terms in a form related to binomial coefficients but encounters difficulties in simplification. Some participants suggest that expanding the terms leads to recognizable patterns, such as sums of squares. Others inquire about clearer explanations and the source of the problem, indicating a need for additional context.

Discussion Status

The discussion is ongoing, with participants providing insights and suggesting evaluations of the sequences P_r and Q_r. There is a request for clearer explanations and further exploration of the terms involved, indicating a collaborative effort to understand the problem better.

Contextual Notes

Participants note that the problem is sourced from a book titled "HIGHER ALGEBRA" by Hall & Knight, which may contain additional context or solutions that are not fully elaborated in the thread. There is also mention of factorial notations in the final result, suggesting complexity in the evaluation.

Vineeth T
Messages
31
Reaction score
0

Homework Statement



If P r=(n-r)(n-r+1)(n-r+2)...(n-r+p-1)
Qr= r(r+1)(r+2)...(r+q-1)
Find P1Q1+P2Q2+...
+Pn-1Qn-1

Homework Equations


The Attempt at a Solution


I tried to bring the general term in the form of a coefficient of x in the binomial expansion.
But it does not simplify to that form.
can anyone give me a better way to approach the problem?
 
Physics news on Phys.org
Its a lot of messing around with, but I'll give you the basic idea, note that when you expand, the integers turn out to be sum of squares, that is (n-1)(n)(2n-1)/6. The rn turn out to be (n-1)rn. You just do this to the different terms to get the final thing, which then you might be able to factor out. Try doing this and post your result, shouldn't be too hard.

Thanks, Bonaparte
 
Bonaparte said:
Its a lot of messing around with, but I'll give you the basic idea, note that when you expand, the integers turn out to be sum of squares, that is (n-1)(n)(2n-1)/6. The rn turn out to be (n-1)rn. You just do this to the different terms to get the final thing, which then you might be able to factor out. Try doing this and post your result, shouldn't be too hard.

Thanks, Bonaparte

Can you explain it more clearly?
Also the source of this problem is from a book called "HIGHER ALGEBRA" by Hall&Knight.
If you have this book see the answer (only the final result is given) in pg:328.Q no:27
The answer even has factorial notations in it.
 
Last edited:
Vineeth T said:
Can you explain it more clearly?
Also the source of this problem is from a book called "HIGHER ALGEBRA" by Hall&Knight.
If you have this book see the answer (only the final result is given) in pg:328.Q no:27
The answer even has factorial notations in it.

Try evaluating P1 , P2 , ... , Pn-1 ...

Then Q1 , Q2 , Q3 , ... , Qn-1...

Don't over simplify...

Then You evaluate P1Q1+P2Q2+...
+Pn-1Qn-1...

First do this much. What do you get ?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
34
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
10
Views
3K