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An alternating squares "aha" puzzle

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Evaluate 1002 - 992 + 982 - 972 + 962 - 952 + ... + 22 - 12 = ?
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5050?

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What about

This could be split into 50 pairs, each given by (2n)^2 - (2n-1)^2 = 4n-1. Summing this between 1 and 50 gives 2n(n+1)-n = 5050. Not an "aha" solution, but I had to write it down

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Posted · Report post

What about

This could be split into 50 pairs, each given by (2n)^2 - (2n-1)^2 = 4n-1. Summing this between 1 and 50 gives 2n(n+1)-n = 5050. Not an "aha" solution, but I had to write it down

Hi Joe, and welcome to the Den.

Pairing is the right first step. Can you make the pairs even simpler?

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There are 50 pairs of n

2 - (n -1)2.
n2 - (n -1)2= n + (n- 1)
So 1002 - 992 + 982 - 972 + 962 - 952 + ... + 22 - 12 can be written
100 + 99 + 98+ ... + 2 + 1 = (100 x 101)/2 = 5050

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I'm not sure it makes it any easier but you could pair from opposite ends, like in the sum to n.

Each pair is then equal to 101 times the difference between the two terms, summing to 101(100-99+98...+2-1) as the terms become negative for higher odd numbers which should give 101(50) =5050

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Good job all. Several great answers.

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Let S

n be a n x n square with it's sides parallel to coordinate axis and bottom left corner at (0,0).
We start with S100, then remove S99, then add S98, then remove S97, etc.
Let's paint green all unit squares of S100 that stay, and let's paint red all unit squares that are removed.
We can see that S100 is divided into 100 L-shaped areas with alternating colors:
most outer area is green, most inner area (degenerated to just 1 unit square) is red.
Out of two neighboring L-shaped areas, the bigger one has two more unit squares than the smaller one.
We have 50 such red-green pairs, therefore we have 100 green unit squares more than red unit squares.
The area of S100 equals 10000, it's half equals 5000, so there has to be 4950 red squares and 5050 green squares.
Number of green squares (5050) is the answer.
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Let S

n be a n x n square with it's sides parallel to coordinate axis and bottom left corner at (0,0).

We start with S100, then remove S99, then add S98, then remove S97, etc.

Let's paint green all unit squares of S100 that stay, and let's paint red all unit squares that are removed.

We can see that S100 is divided into 100 L-shaped areas with alternating colors:

most outer area is green, most inner area (degenerated to just 1 unit square) is red.

Out of two neighboring L-shaped areas, the bigger one has two more unit squares than the smaller one.

We have 50 such red-green pairs, therefore we have 100 green unit squares more than red unit squares.

The area of S100 equals 10000, it's half equals 5000, so there has to be 4950 red squares and 5050 green squares.

Number of green squares (5050) is the answer.

Nice.

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