Problem:
If Tn=1+2+3+⋯+n and
Pn=T2−1T2⋅T3−1T3⋅T4−1T4⋅⋯⋅Tn−1Tn for n=2,3,4,…,
then P1991 is closest to which of the following numbers?
Answer Choices:
A. 2
B. 2.3
C. 2.6
D. 2.9
E. 3.2
Solution:
Since Tn=2(n+1)n,Pn=2n(n+1)−22n(n+1)Pn−1=(n+2)(n−1)(n+1)nPn−1.
Therefore
P1991=1993⋅19901992⋅1991⋅(1992⋅19891991⋅1990P1989)=19931991⋅19891991P1989=19931991⋅19891991⋅(1991⋅19881990⋅1989P1988)=19931991⋅19881990P1988=⋯=19931991⋅kk+2Pk=⋯=19931991⋅24P2=19931991⋅3
so 2.9 is closest to P1991.
OR
Note that
Pn=k=2∏n(k+2)(k−1)k(k+1)=(∏k=2n(k+2))(∏k=2n(k−1))(∏k=2nk)(∏k=2n(k+1))=(2⋅3(n+2)!)(n−1)!n!(2(n+1)!)=n+23n
so P1991=19933⋅1991 which is closest to 2.9.