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1+2+....+(n-1)+n+(n-1)+...+2+1 = 2( 1+2+...(n-1)) +n = 2((1+(n-1))*(n-1)/2 +n = n(n-1) +n = n^2
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This works in any numerical system to a base B with sequences of 1's up to a length of B-1! Example: Let B=3. 11²=121 (In decimal: (1x3+1x1)²=16=1x3²+2x3+1x1)
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palindromic?
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It's palindromic for sequences of 1's up to a length of B-1, where B is the base of the numerical system (e.g. decimal system => base is 10 => up to 9 1's)
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Nothing unusual here, one don't even need to apply a formula for triangular numbers here. All numbers cancel out and we're left with n times n aka n². For example n:=3 1+2+3+3-1+3-2=3²pic.twitter.com/f7tpEqGM7m
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Pascal's Triangle ¯\_(ツ)_/¯
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There are ten items in last row. Must be nine ;)
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Isn't that a Pascal triangle?
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