1991 AHSME Problems/Problem 25

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Problem

If $T_n=1+2+3+\cdots +n$ and \[P_n=\frac{T_2}{T_2-1}\cdot\frac{T_3}{T_3-1}\cdot\frac{T_4}{T_4-1}\cdot\cdots\cdot\frac{T_n}{T_n-1}\] for $n=2,3,4,\cdots,$ then $P_{1991}$ is closest to which of the following numbers?

$\text{(A) } 2.0\quad \text{(B) } 2.3\quad \text{(C) } 2.6\quad \text{(D) } 2.9\quad \text{(E) } 3.2$

Solution

$\fbox{D}$ It is well known that $T_n = \frac{n(n+1)}{2}$, so we can put this in and simplify to get that the general term of the product is $\frac{n(n+1)}{(n+2)(n-1)}.$ Now by writing out the terms, we see that almost everything cancels (telescopes), giving $\frac{3 \times 1991}{1993}$ which is almost $3$, so closest to $2.9.$

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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