Math Problem

Okay, since there are no takers heres a clue:

Since the series consists of n terms all involving n squared, try solving the general cubic equation

nth term = a.n^3 + b.n^2+c.n +d

and instead of the 100th term give me the 10th (without cheating).

What - still no takers ?

Heres the solution:

The nth term =(2n^3 +3n^2 + n)/6

so the 10th term = (2000+300+10)/6 = 385

good example of me, myself and I

Re: Math Problem

Just curious, in what context did you give this homework?

Gouki

My apologies if you were working on it. ( There's nothing to stop you proving the answer.) In my experience of other people's math postings on a2k they are left hanging in the air for too long. Perhaps you could offer one of your own for entertainment.



Satt

I manage a small Prep School in the UK. What I thought was a similar example was given on an examination paper for eleven year old entry into Manchester Grammar School, however that one only required the generation of a quadratic or a bit of trial and error with the mapping rule. (Predict the number of diagonals drawn in successive cyclic polygons) My example concerned the number of points needed to build a growing square based pyramid.

fresco..
Related problems will be interesting:

S = 2*2+4*4+6*6+..+(2n)*(2n)

S = 1*1*1+2*2*2+3*3*3+4*4*4 +..+ n*n*n

..

Satt

At first glance (10 minute break) the answer to the first looks like four times the answer to the original problem. The answer to the second involves a general quartic in n, or the solution of the 5 simultaneous equations for the constants a,b,c,d,e.

The coefficient as a constant term ('e') must be zero.

Agreed !