1, solution: Let the number of bricks in the Temple of Heaven * * * be A3n, give A 1=9, d=9, and find S3n. \r\n2, first floor: A 1, A2,? , one. \r\n3, second floor: A(n+ 1), A(n+2),? , A(n+n).\r\n4, the third floor: A(2n+ 1), A(2n+2),? , A(2n+n).\r\n5, then the first item of the third floor, A(2n+ 1)=a 1+2nd, is nd more than the first item of the second floor, A(n+ 1)=a 1+nd. \r\n6, then n 2d = 729, and n=√(729/9)=9. \r\n7, then s3n = s27 = na1+n (n-1) d/2 = 27× 9+27× (27-1)× 9/2 = 3402. \ r \ For detailed information about how to write the Temple of Heaven, please visit:/ask/ebbe3161582 9921.html? Zd view more content