Create a chart on a piece of paper like the one below. It could be any NxN square; there’s nothing special about the fact that N=7 in this example.

1 | 2 | 3 | 4 | 5 | 6 | 7 |

8 | 9 | 10 | 11 | 12 | 13 | 14 |

15 | 16 | 17 | 18 | 19 | 20 | 21 |

22 | 23 | 24 | 25 | 26 | 27 | 28 |

29 | 30 | 31 | 32 | 33 | 34 | 35 |

36 | 37 | 38 | 39 | 40 | 41 | 42 |

43 | 44 | 45 | 46 | 47 | 48 | 49 |

Give the chart to a Member of the Audience (or a student in your class, or your Great-Aunt Mabel, or…). Without you watching, this person should circle a number, cross off all the other numbers in that row and column, and then pass the paper to another person. The next person circles one of the remaining numbers, crosses off everything else in that number’s row and column, and passes it along. This continues until everything on the chart has either been circled or crossed off. At this point, the last person to circle a number adds up all the circled numbers and think very hard about the sum.

You, meanwhile, know that people are circling numbers but don’t know which numbers are being circled. Once the final person has added up the numbers and telepathically thinks of the sum, you will use your Amazing Mindreading Powers to announce the sum!

Which will be 175.

Or, in general for an NxN square, will be (N^{3}+N)/2. Pretty cool, isn’t it? Of course, if you do this trick more than once for the same crowd, you might want to vary the size of the square to mask the fact that the same size square always leads to the same answer.

Here’s a brief justification for the total. Because numbers in the same row and column as a circled number are crossed out, it turns out exactly one number in each row is circled and, at the same time, exactly one number in each column is circled.

When you add, think of adding the numbers row by row.

- The first number will be between 1 and 7 (inclusive). Call it A, where 1≤A≤7.
- The second number will be between 8 and 14 (inclusive). Call it 7+B, where 1≤B≤7.
- The third number will be between 15 and 21 (inclusive). Call it 2·7+C, where 1≤C≤7.
- Continue all the way to the seventh number, which will be between 43 and 49. Call it 6·7+G, where 1≤G≤7.

When you add these all up you get (A)+(7+B)+(2·7+C)+…+(6·7+G). Reordering gives (7+2·7+…+6·7)+(A+B+…+G), which can be simplified as (1+2+…+6)·7+(A+B+…+G).

Because you don’t know which numbers were circled, you don’t know what A, B, etc. are. But you do know that one number was picked in each column! Each number in the first column is 1 more than a multiple of 7, each number in the second column is 2 more than a multiple of 7, etc.

Since one number is picked out of each column, the numbers A, B, …, G are 1, 2, …, 7 in some order. This means that A+B+…+G is just 1+2+…+7, and the total amount is (1+2+…+6)·7+(1+2+…+7). In general, for an NxN square, the total will be (1+2+…+[N-1])·N+(1+2+…+N). The first summation is (N-1)N/2 and the second summation is N(N+1)/2, so the total becomes {(N-1)N/2}·N+N(N+1)/2. Factoring out N/2 gives (N/2)(N^{2}-N+(N+1)), which becomes N(N^{2}+1)/2 or (N^{3}+N)/2.

But of course, you don’t have to give away your secret.

*I found this trick on curiousmath.*