lo g 2 ( n ! ) 1 + lo g 3 ( n ! ) 1 + lo g 4 ( n ! ) 1 + … + lo g n ( n ! ) 1 = ?
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lo g 2 ( n ! ) 1 + lo g 3 ( n ! ) 1 + lo g 4 ( n ! ) 1 + . . . + lo g n ( n ! ) 1
= lo g 2 lo g ( n ! ) 1 + lo g 3 lo g ( n ! ) 1 + lo g 4 lo g ( n ! ) 1 + . . . + lo g n lo g ( n ! ) 1
= lo g n ! lo g 2 + lo g n ! lo g 3 + lo g n ! lo g 4 + . . . + lo g n ! lo g n
= lo g n ! lo g 2 + lo g 3 + lo g 4 + . . . + lo g n
= lo g n ! lo g 2 ⋅ 3 ⋅ 4 ⋅ . . . ⋅ n
= lo g n ! lo g n !
= 1
Great solution @David Vreken !
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Thank you! I like yours better as it was a little bit quicker, but I wanted to show a slightly different way.
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Thanks!, this problem is actually inspired by 3b1b's lockdown math ep. 6. He shows a similar question and solves just like your method. Here is the link
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Property References:
1 . lo g ( a b ) = lo g ( a ) + lo g ( b ) 2 . lo g a b = lo g b a 1
Evaluating the expression
= lo g 2 ( n ! ) 1 + lo g 3 ( n ! ) 1 + lo g 4 ( n ! ) 1 + … + lo g n ( n ! ) 1 = lo g n ! ( 2 ) + lo g n ! ( 3 ) + lo g n ! ( 4 ) + … + lo g n ! ( n ) [ 2 ] = lo g n ! ( 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋯ n ) [ 1 ] = lo g n ! ( n ! ) = 1