
Re: [mathlearn] can any one solve my challenge. lets see.
Posted:
Dec 26, 2010 2:52 PM



On Sun, 26 Dec 2010, tcymark@yahoo.com wrote:
> what is the 88th term to this a.p. 4, 8, 12, 16......
I couldn't figure out what an "a.p." was, so I chose the nearest convenient formula that produces these as the first four terms, and found:
If A(n) = 4n + (n1)(n2)(n3)(n4), I sure enough get A(1) = 4*1 + 0*(1)*(2)*(3), or 4*1; A(2) = 4*2 + 1*(0)*(1)*(2), or 4*2; A(3) = 4*3 + 2*1*0*(1), or 4*3; and A(4) = 4*4 + 3*2*1*0, or 4*4.
So we get the first four terms correct using this formula. Then
A(88) = 4*88 + 87*86*85*84, or 53421832.
Ralph A. Raimi Tel. 585 275 4429 or (home) 585 244 9368 Dept. of Mathematics, Univ. of Rochester, Rochester, NY 14627 <http://www.math.rochester.edu/people/faculty/rarm/>

