Telescoping Charm!

Algebra Level 4

If T n = 1 + 2 + 3 + + n T_{n} = 1 + 2 + 3 + \dots + n , then P n = T 2 T 2 1 T 3 T 3 1 T 4 T 4 1 T n T n 1 P_n = \frac{T_2}{T_2 - 1} \cdot \frac{T_3}{T_3 - 1} \cdot \frac{T_4}{T_4 - 1} \dotsm \frac{T_n}{T_n - 1} for n = 2 , 3 , 4 , n = 2, 3, 4, \dots , then P 1991 P_{1991} is closest to what positive integer?


The answer is 3.

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1 solution

Revanth Gumpu
Aug 20, 2015

T n = 1 + 2 + 3 + + n = n ( n + 1 ) 2 T_{n} = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2} T n 1 = n ( n + 1 ) 2 1 = n 2 + n 2 2 = ( n 1 ) ( n + 2 ) 2 \Rightarrow\ T_n - 1 = \frac{n(n + 1)}{2} - 1 = \frac{n^2 + n - 2}{2} = \frac{(n - 1)(n + 2)}{2} T n T n 1 = n ( n + 1 ) / 2 ( n 1 ) ( n + 2 ) / 2 = n ( n + 1 ) ( n 1 ) ( n + 2 ) , \Rightarrow\ \frac{T_n}{T_n - 1} = \frac{n(n + 1)/2}{(n - 1)(n + 2)/2} = \frac{n(n + 1)}{(n - 1)(n + 2)},

P 1991 = T 2 T 2 1 T 3 T 3 1 T 4 T 4 1 T 1991 T 1991 1 \Rightarrow\ P_{1991} = \frac{T_2}{T_2 - 1} \cdot \frac{T_3}{T_3 - 1} \cdot \frac{T_4}{T_4 - 1} \dotsm \frac{T_{1991}}{T_{1991} - 1}

2 3 1 4 3 4 2 5 4 5 3 6 1990 1991 1989 1992 1991 1992 1990 1993 \Rightarrow \frac{2 \cdot 3}{1 \cdot 4} \cdot \frac{3 \cdot 4}{2 \cdot 5} \cdot \frac{4 \cdot 5}{3 \cdot 6} \dotsm \frac{1990 \cdot 1991}{1989 \cdot 1992} \cdot \frac{1991 \cdot 1992}{1990 \cdot 1993}

3 1991 1993 \Rightarrow \frac{3 \cdot 1991}{1993}

3 \Rightarrow \approx \boxed{3}

Moderator note:

Simple standard approach using the telescoping product.

Great question! Wonderful solution as well :)

John Frank - 5 years, 3 months ago

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Thank you!

Revanth Gumpu - 5 years, 3 months ago

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