If T_n=1+2+3+…+n and P_n&=(T_2)/(T_2-1) ×(T_3)/(T_3-1) & ×(T_4)/(T_4-1) …(T_n)/(T_n-1) for n=2, 3, 4, …, then P_1991 is closest to which of the following numbers?

1991 AMC 12 Problem 25

Attempt Problem 25 of the 1991 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1991 AMC 12 solutions, or check the answer key.

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25.

T_n=1+2+3+\cdots+n

and

\begin{aligned} P_n&=\frac{T_2}{T_2-1} \cdot\frac{T_3}{T_3-1}\\ &\quad\cdot\frac{T_4}{T_4-1} \cdots\frac{T_n}{T_n-1} \end{aligned}

for

n=2,

3,

4,

\ldots,

then

P_{1991}

is closest to which of the following numbers?

2.0

2.3

2.6

2.9

3.2

Factor

T_k-1

and simplify each factor

\frac{T_k}{T_k-1}

Write each factor as

\frac{k}{k-1}\cdot\frac{k+1}{k+2}

and telescope

Since

T_k=\frac{k(k+1)}{2}

and

T_k-1=\frac{(k-1)(k+2)}{2},

\frac{T_k}{T_k-1} =\frac{k}{k-1}\cdot\frac{k+1}{k+2}.

The product telescopes:

\begin{aligned} P_n &=\left(\prod_{k=2}^n\frac{k}{k-1}\right) \left(\prod_{k=2}^n\frac{k+1}{k+2}\right)\\ &=n\cdot\frac3{n+2}\\ &=\frac{3n}{n+2}. \end{aligned}

For

n=1991,

this is just under

3

and is closest to

2.9.

Thus the correct answer is D.