Challenging Mathematical Problems with Elementary Solutions, 第 1 卷Holden-Day, 1964 - 440 頁 |
搜尋書籍內容
第 1 到 3 筆結果,共 66 筆
第 62 頁
... equal to m / 2 , and if m is odd , both sides equal ( m - 1 ) / 2 . S This fact can be used to simplify the expression for S. We have n - 3 = ( 플 ++ ) + ( = ++ ) 4 n 3 3 2 n 6 3 ( -1 ) - + 2 + + 4 4 4 + + ( 1 ++ ) Since there are q + 1 ...
... equal to m / 2 , and if m is odd , both sides equal ( m - 1 ) / 2 . S This fact can be used to simplify the expression for S. We have n - 3 = ( 플 ++ ) + ( = ++ ) 4 n 3 3 2 n 6 3 ( -1 ) - + 2 + + 4 4 4 + + ( 1 ++ ) Since there are q + 1 ...
第 188 頁
... equal to Gn - 1 + G1Gn - 2 + G2Gn - 3 + ··· + Gn - 2G1 + Gn - 1 = = ( the first and last terms correspond to the cases l 1 and 7 : n ; these terms are obviously equal to Gn - 1 ) . If the second vertex of the triangle containing A , is ...
... equal to Gn - 1 + G1Gn - 2 + G2Gn - 3 + ··· + Gn - 2G1 + Gn - 1 = = ( the first and last terms correspond to the cases l 1 and 7 : n ; these terms are obviously equal to Gn - 1 ) . If the second vertex of the triangle containing A , is ...
第 212 頁
... equal to the length of that interval ( recall that in our case OA = OC 1 ) . Therefore the probability that a point ( x , y ) lies within any rectangle inside OABC is equal to the area of that rectangle . Since any region can be covered ...
... equal to the length of that interval ( recall that in our case OA = OC 1 ) . Therefore the probability that a point ( x , y ) lies within any rectangle inside OABC is equal to the area of that rectangle . Since any region can be covered ...
其他版本 - 查看全部
常見字詞
A₁ A₂ An+m arrangements b₁ B₂ binomial coefficients binomial theorem bishops black squares C₁ chessboard chord circle coefficient color column compute the number Consequently consider corresponding customers denote determine the number diagonals digits dihedral angle divided divisible draw equally likely possible equation equidistant equivalence classes exactly example experiment favorable outcomes follows formula given Hence inclusion and exclusion intersection k-gons knights L₁ length mathematical induction maximum number n-gon number of different number of favorable number of paths number of shortest obtain pairs partition passengers plane points A1 polygons positive integers possible outcomes Pr{E probability theory problem 54 prove queens rectangle relatively prime remaining required probability rooks segment selected at random sequence shortest paths side solution to problem solved sphere square controlled Suppose T₂ total number triangle unfavorable values vertex vertices