Adv Alegbra HW Solutions 6

Adv Alegbra HW Solutions 6 - 1 ) n X i = i = n X i = i 2 +...

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6 Solution. As indicated in Figure 1.3, a rectangle with height n + 1 and base n i = 1 i k can be subdivided so that the shaded staircase has area n i = 1 i k + 1 , while the area above it is 1 k + ( 1 k + 2 k ) + ( 1 k + 2 k + 3 k ) +···+ ( 1 k + 2 k +···+ n k ). One can prove this, for f xed k , by induction on n 1. 1 2 3 5 4 k k k k k 1 5 4 + k k k 1 2 3 5 4 k +1 k +1 k +1 k +1 k +1 1 + 2 3 4 k k k k 3 k 1 + 2 k k 1 + 2 + + k k 1 + 2 + + k k k 3 + Figure 1.3 Alhazan s Dissection (iii) Given the formula n i = 1 i = 1 2 n ( n + 1 ) , use part (ii) to derive the formula for n i = 1 i 2 . Solution. ( n +
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Unformatted text preview: 1 ) n X i = i = n X i = i 2 + n X i = i X p = p = n X i = i 2 + n X i = 1 2 i ( i + 1 ) = n X i = i 2 + 1 2 n X i = i 2 + 1 2 n X i = i . Therefore, ( n + 1 1 2 ) n X i = i = 3 2 n X i = i 2 , and so n X i = i 2 = 2 3 ( n + 1 2 ) 1 2 n ( n + 1 ) = 1 3 1 2 ( 2 n + 1 ) n ( n + 1 ) = 1 6 ( 2 n + 1 ) n ( n + 1 )....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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