Unit 2 · Topic 14 · Number Play
Pixel asked for the sum of the first 5 odd numbers: 1+3+5+7+9.
Kabir reached for his fingers to add them one at a time.
Anaya said '25' immediately, without adding a single pair.
"Add the first 5 odd numbers," said Pixel. "1, 3, 5, 7, 9." Kabir started counting on his fingers, adding one at a time. Anaya answered before he'd added the second term. "25."
"How?" asked Kabir, once he'd caught up and confirmed 1+3+5+7+9 really was 25. "It's always a square number," said Anaya. "There are 5 odd numbers here, so the answer is 5 squared, 25."
Pixel tried a bigger one: the first 8 odd numbers, 1+3+5+7+9+11+13+15. "8 terms," said Anaya, "so 8 squared, 64." Kabir checked by adding all eight, and got exactly 64.
"Why does that work?" Kabir asked. Pixel drew it with dots: one dot, then an L-shaped layer of 3 more dots around it making a 2x2 square, then an L-shaped layer of 5 more making a 3x3 square, then 7 more making a 4x4 square. "Each new odd-numbered layer of dots grows the square by exactly one row and one column — that's exactly why the running total is always a perfect square."
"So the rule is," said Anaya, "the sum of the first n odd numbers is always n squared. No pairing needed this time — you just count how many odd numbers there are, and square that count." Kabir tested it on the first 10 odd numbers: predicted 100, checked by adding, and got exactly 100.
The sum of the first n odd numbers (1, 3, 5, ..., up to the nth odd number) always equals n squared (n x n). For example, the first 5 odd numbers sum to 5 x 5 = 25.
This can be seen visually: each new odd number added is exactly the right size to grow a square of dots by one more row and one more column, so the running total is always a perfect square.
Unlike the natural-number or even-number sums, no pairing step is needed here — simply count how many odd numbers are being added (that count is n) and square it.
Add the first 6 odd numbers, then confirm the total equals 6 squared.
Predict the sum of the first 12 odd numbers without adding, then check by adding them all.
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