Posted October 12, 2013 Evaluate 100^{2 }- 99^{2} + 98^{2} - 97^{2} + 96^{2} - 95^{2} + ... + 2^{2} - 1^{2} = ? 0 Share this post Link to post Share on other sites

0 Posted October 12, 2013 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 0 Share this post Link to post Share on other sites

0 Posted October 12, 2013 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 downHi Joe, and welcome to the Den. Pairing is the right first step. Can you make the pairs even simpler? 0 Share this post Link to post Share on other sites

0 Posted October 12, 2013 There are 50 pairs of n^{2} - (n -1)^{2}. n^{2} - (n -1)^{2}= n + (n- 1) So 100^{2} - 99^{2} + 98^{2} - 97^{2} + 96^{2} - 95^{2} + ... + 2^{2} - 1^{2} can be written 100 + 99 + 98+ ... + 2 + 1 = (100 x 101)/2 = 5050 0 Share this post Link to post Share on other sites

0 Posted October 12, 2013 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 0 Share this post Link to post Share on other sites

0 Posted October 12, 2013 Good job all. Several great answers. 0 Share this post Link to post Share on other sites

0 Posted October 14, 2013 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 S_{100}, then remove S_{99}, then add S_{98}, then remove S_{97}, etc. Let's paint green all unit squares of S_{100} that stay, and let's paint red all unit squares that are removed. We can see that S_{100} 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 S_{100} 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. 0 Share this post Link to post Share on other sites

0 Posted October 15, 2013 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 S_{100}, then remove S_{99}, then add S_{98}, then remove S_{97}, etc. Let's paint green all unit squares of S_{100} that stay, and let's paint red all unit squares that are removed. We can see that S_{100} 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 S_{100} 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. 0 Share this post Link to post Share on other sites

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^{2 }- 99^{2}+ 98^{2}- 97^{2}+ 96^{2}- 95^{2}+ ... + 2^{2}- 1^{2}= ?## Share this post

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