Statement The identity 4 Pyrn=Tet2n4\,\mathrm{Pyr}_n=\mathrm{Tet}_{2n}4Pyrn=Tet2n holds, where Pyrn=n(n+1)(2n+1)6\mathrm{Pyr}_n=\dfrac{n(n+1)(2n+1)}{6}Pyrn=6n(n+1)(2n+1) and Tetm=m(m+1)(m+2)6\mathrm{Tet}_m=\dfrac{m(m+1)(m+2)}{6}Tetm=6m(m+1)(m+2). Verify it for n=4n=4n=4. Solution Pyr4=4⋅5⋅96=30\mathrm{Pyr}_4=\dfrac{4\cdot 5\cdot 9}{6}=30Pyr4=64⋅5⋅9=30, so 4 Pyr4=4⋅30=120.4\,\mathrm{Pyr}_4=4\cdot 30=\boxed{120}.4Pyr4=4⋅30=120. On the other hand Tet8=8⋅9⋅106=120\mathrm{Tet}_8=\dfrac{8\cdot 9\cdot 10}{6}=120Tet8=68⋅9⋅10=120, matching the identity.