Note that \ (c = a \) when \ (| ac | = 0 \), and \ (| xy | = | yx | \) so \ (\ displaystyle \ sum \

2 # Xinhuanet published in 2014-6-9 02:34 PM Show author to consider a number of lines 1, 2, I first removed bishop mcdevitt wyncote after a sigma, ..., m, and the distance to see: become \ ( \ sum \ limits_ {a = 1} ^ {m} {\ left (\ left | a-1 \ right | + \ left | a-2 \ right | + ... + \ left | am \ right | \ right )} = \ sum \ limits_ {k = 1} ^ {m} {\ left [\ left (1 + 2 + ... + \ left (mk \ right) \ right) + \ left (1 + 2 +. .. + \ left (k-1 \ right) \ right) \ right]} \) in simplified into \ (\ frac {1} {2} \ left (\ sum \ limits_ {k = 1} ^ {m } {\ left ({{m} ^ {2}} + m \ right) +} 2 \ sum \ limits_ {k = 1} ^ {m} {{{k} ^ {2}}} - 2 \ left (m + 1 \ right) \ sum \ limits_ {k = 1} ^ {m} {k} \ right) \) asked for simplification finished just do not know if that is there a faster way to be completed [this post Finally hua0127 at 2014-6-9 02:36 PM edited by]
Note that \ (c = a \) when \ (| ac | = 0 \), and \ (| xy | = | yx | \) so \ (\ displaystyle \ sum \ limits _ {a = 1} ^ {m} \ sum \ limits _ {c = 1} ^ {m} | ca | = 2 \ sum \ limits _ {a = 1} ^ {m-1} \ sum \ limits _ {c = a + 1} ^ {m } | ca | \) so that \ (p = ca, q = m-a + 1 \), then \ (\ displaystyle 2 \ sum \ limits _ {a = 1} ^ {m-1} \ sum \ limits _ {c = a + 1} ^ {m} | ca | = 2 \ sum \ limits _ {a = 1} ^ {m-1} \ sum \ limits _ {p = 1} ^ {ma} p = 2 \ sum \ limits _ {a = 1} ^ {m-1} C_ {2} ^ {m-a + 1} = 2 \ sum \ limits _ {q = 2} ^ {m} C_ {2} ^ {q } = 2C_ {3} ^ {m + 1} = \ frac {m ^ {3} -m} {3} \) upstream, fourth equal sign with the Pascal's theorem as Friend Offline
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