Q&A for How to Add a Sequence of Consecutive Odd Numbers

Whenever you add together two consecutive numbers, the sum is an odd number. When you divide any odd number by two, you get a quotient beginning with the first digit or digits you started with and ending in .5. Then if you subtract the first digit(s) from the quotient, you're left with .5. This is true when working with any consecutive numbers, two-digit or otherwise.

To add say 21+23+...91, just pretend it started at 1 and subtract off the excess. First use the formula on 1 + ...91 then subtract the part you wanted to leave out 1+ ...+ 19 which you can sum with the same formula.

The sum of the last 5 numbers in the sequence make up 495. If we state that "n" represents any odd number and start our sequence: n + (n+2) + (n+4) + (n+6) + (n+8) = 425, where n = 81. If we continued this sequence, the last 5 consecutive odd numbers can be represented by: (n+14) + (n+16) + (n+18) + (n+20) + (n+22). By substituting n = 81 into this sequence, we get 495. Another way is to write out 12 consecutive odd numbers starting with 81 and then adding the last 5 numbers.

72 consecutive odd numbers make up 5184 starting with 1: 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 129 131 133 135 137 139 141 143

For the odds: 1, 3, 5, 7, 9, 11, 13, 15, 17; For the evens: 0, 2, 4, 6, 8, 10, 12, 14, 16; For both: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17.

To solve this equation, you must divide both sides by 2. 126 divided by 2 equals 63. So, yes, this is correct.

The formula is simply n².

2n - 1, where n is the sequential position. For example, in the sequence 1, 3, 5, 7..., 1 is in sequence position #1, so that 2n - 1 = 2(1) - 1 = 1. 3 is in sequence position #2, so that 2n - 1 = 2(2) - 1 = 3, and so on.

Let x be the first number. Then x+2 is the second (middle) number, and x+4 is the third number. Their sum is x + (x+2) + (x+4) = 3x + 6 = (x + 2)² [the square of the middle number] = x² + 4x + 4. Subtract 3x from both sides, and subtract 6 from both sides: 0 = x² + x - 2 = (x+2)(x-1), so that x = -2, 1. Reject (-2) as an answer, because it's not an odd number, leaving x = 1. That means the three consecutive odd numbers are 1, 3, and 5. (Their sum is 9, which is the square of the middle number.)

If you're asking about the sum of the first 1,000 odd numbers, it's 1,000² = 1,000,000 (one million).

"n" refers to the number of numbers in the sequence. (2n + 1) refers to the last (largest) number in the sequence. If you know n or (2n + 1), use them as shown in Part 1 above.

No. The formula for finding the sum of the first n consecutive even numbers (from 2 to n) is [n(n+2)] / 4. For example, the sum of the even numbers from 2 to 12 is [(12)(14)] / 4 = 42.

2n + 2 = 10. Subtract 2 from both sides: 2n = 8. Divide both sides by 2: n = 4.

The 81st number in the sequence is (81 - 1)(2) + 1 = 161, which is the "ending point." So the sum of the sequence is [(161 + 1) / 2]² = 81² = 6561.

There is apparently no such formula. However, there is a fairly easy way to accomplish what you want. Use the above formula to find the sum of all the odd numbers from (to use your example) 1 through 91. Then use the same formula to find the sum of all the odd numbers from 1 through 17. Then subtract the second sum from the first sum.