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Dulce

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Summing Series: n.1+(n-1).2+(n-2).3+...+n.1?

How would you find the sum of this series:

n.1+(n-1).2+(n-2).3+...+n.1

I know that the answer to it is [n(n+1)(n+2)]/6 But neither me or the teacher could figure out how to solve it.

Thank you.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Summing Series: n.1+(n-1).2+(n-2).3+...+n.1?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Summing Series: n.1+(n-1).2+(n-2).3+...+n.1?

Here is a practical application of the formula.

There are r points (r ≥ 4) equally spaced out on a circle, and four points A, B, C, D are chosen randomly. Prove that the probability that the chord AB intersects the chord CD is

.

Proof:

Suppose there are k points on the arc between A and B (not including A and B) going clockwise from A to B. If C is one of these points, then D must be one of the r−2−k points on the arc going anticlockwise from A to B. Similarly if C and D are interchanged. Hence there are

ways of choosing chord CD given this configuration of A and B.

Since k can range from 1 to r−3, the total number of ways of choosing intersecting chords is

which, by taking

in the formula we have proved, is

But

is the total number of ways of choosing the points A, B, C, D randomly. (A can be any point (doesn’t matter which); then there are r−1 choices for B, and after that r−2 choices for C, and after that r−3 choices for D.) Hence the required probability is

.

Last edited by JaneFairfax (2008-11-14 18:00:02)