Finding the sum P1Q1 + P2Q2 + ... + Pn-1Qn-1 with products of consecutive integers

[Vineeth T]

Homework Statement

If P _r=(n-r)(n-r+1)(n-r+2)...(n-r+p-1)
Q_r= r(r+1)(r+2)...(r+q-1)
Find P₁Q₁+P₂Q₂+...
+P_n-1Q_n-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?

 

[Bonaparte]

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

 

[Vineeth T]

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.

 

[sankalpmittal]

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 , ... , P_n-1 ...

Then Q₁ , Q₂ , Q₃ , ... , Q_n-1...

Don't over simplify...

Then You evaluate P₁Q₁+P₂Q₂+...
+P_n-1Q_n-1...

First do this much. What do you get ?