Question

H-917 Proposed by Benjamin Lee Warren, New York, NY Let $O_n = \frac{1}{3}n(2n^2 + 1)$ denote the nth Octahedral number and $T_n = \frac{1}{6}n(n+1)(n+2)$ denote the nth Tetrahedral number. Prove the identity $O_{F_{2n}} + T_{F_{2n-1}-1} = T_{F_{2n+1}-1}$.

          H-917 Proposed by Benjamin Lee Warren, New York, NY
Let $O_n = \frac{1}{3}n(2n^2 + 1)$ denote the nth Octahedral number and $T_n = \frac{1}{6}n(n+1)(n+2)$ denote
the nth Tetrahedral number. Prove the identity
$O_{F_{2n}} + T_{F_{2n-1}-1} = T_{F_{2n+1}-1}$.

H-917 Proposed by Benjamin Lee Warren, New York, NY
Let On = (1)/(3)n(2n^2 + 1) denote the nth Octahedral number and Tn = (1)/(6)n(n+1)(n+2) denote
the nth Tetrahedral number. Prove the identity
OF2n + TF2n-1-1 = TF2n+1-1.

Added by Jennifer B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Lottie Adams

07/06/2021

Step 1

First, we can simplify the given identity by removing the -1 from both sides, which gives us: OF2n + TFn-1 = TF2n + 1 Now, let's substitute the definitions of Octahedral and Tetrahedral numbers into the equation: (2n)^2 + 2n + 1 + n(n + 1)(n + 2) = 2n(2n + 1)(2n Show more…

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H-917 Proposed by Benjamin Lee Warren, New York, NY Let On = n^2 + n + 1 denote the nth Octahedral number and Tn = n(n + 1)(n + 2) denote the nth Tetrahedral number. Prove the identity: OF2n + TFn-1 - 1 = TF2n + 1 - 1