Unit 2 · Topic 13 · Number Play
Pixel asked for the sum of the first 6 even numbers: 2+4+6+8+10+12.
Kabir started adding left to right, term after term.
Anaya multiplied two numbers together and was done.
"Add the first 6 even numbers," said Pixel. "2, 4, 6, 8, 10, 12." Kabir began: 2+4=6, +6=12, +8=20, +10=30, +12=42.
Anaya had a shortcut ready. "Every even number is just 2 times a counting number," she said. "2=2x1, 4=2x2, 6=2x3, and so on up to 12=2x6. So the whole sum is really 2x(1+2+3+4+5+6)." She used the natural-number sum formula from before: 1+2+3+4+5+6 = 6x7/2 = 21. "So the answer is 2 x 21 = 42."
It matched Kabir's answer exactly. "There's an even faster way, though," said Anaya. "Just like before, pair the ends: 2+12=14, 4+10=14, 6+8=14 — three pairs, each worth 14. 3x14=42."
Pixel pointed out the pattern directly: "For the first n even numbers, the sum is always n times (n+1). Here n=6, so 6x7=42." Kabir tried it on a new case: the first 10 even numbers. "n=10, so 10x11=110." He checked by adding 2+4+6+8+10+12+14+16+18+20, and got exactly 110.
"So there are really two ways to see it," said Anaya. "Double the sum of the first n counting numbers, or use the direct formula n(n+1). Same answer either way — because doubling n(n+1)/2 just gives you n(n+1) again."
The sum of the first n even numbers (2, 4, 6, ..., 2n) equals n x (n+1). For example, the first 6 even numbers sum to 6 x 7 = 42.
This formula comes from the fact that every even number is 2 times a counting number, so the sum of the first n even numbers is exactly double the sum of the first n counting numbers: 2 x [n(n+1)/2] = n(n+1).
The same pairing trick used for consecutive natural numbers also works here: pairing the smallest even number with the largest gives a constant pair-sum, repeated across all the pairs.
Use the formula to find the sum of the first 8 even numbers, then check by adding them one by one.
Find the sum of the first 15 even numbers using only the formula.
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