Home Competitions AHSME 1991 Problem 25
If T n = 1 + 2 + 3 + ⋯ + n T_n=1+2+3+\cdots +n T n = 1 + 2 + 3 + ⋯ + n 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}\cdot\cdots\cdot\frac{T_n}{T_n-1} P n = T 2 − 1 T 2 ⋅ T 3 − 1 T 3 ⋅ T 4 − 1 T 4 ⋅ ⋯ ⋅ T n − 1 T n n = 2 , 3 , 4 , ⋯ , n=2,3,4,\cdots, n = 2 , 3 , 4 , ⋯ , P 1991 P_{1991} P 1991 (A) 2.0 (B) 2.3 (C) 2.6 (D) 2.9 (E) 3.2 \text{(A) } 2.0 \text{(B) } 2.3 \text{(C) } 2.6 \text{(D) } 2.9 \text{(E) } 3.2 (A) 2.0 (B) 2.3 (C) 2.6 (D) 2.9 (E) 3.2
Dataset text NuminaMath-1.5 . Quoted with source.
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