This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 147 Assignment #1 Solutions 1) Prove that: a) 1 2 + 2 2 + Â·Â·Â· + n 2 = n ( n + 1)(2 n + 1) 6 for each n âˆˆ N . Solution: (5 marks). Let P ( n ) be the statement that 1 2 + 2 2 + Â·Â·Â· + n 2 = n ( n + 1)(2 n + 1) 6 . We first show that P (1) holds. But this is true since 1 = 1(1 + 1)(2 Â· 1 + 1) 6 = 6 . We now assume that P ( k ) holds. Then 1 2 + 2 2 + Â·Â·Â· + ( k + 1) 2 = (1 2 + 2 2 + Â·Â·Â· + k 2 ) + ( k + 1) 2 . Since P ( k ) holds, we get that 1 2 + 2 2 + Â·Â·Â· + ( k + 1) 2 = (1 2 + 2 2 + Â·Â·Â· + k 2 ) + ( k + 1) 2 = k ( k + 1)(2 k + 1) 6 + k + 1 2 = ( k + 1)( k (2 k + 1) + 6( k + 1)) 6 = ( k + 1)(2 k 2 + k + 6 k + 6) 6 = ( k + 1)(2 k 2 + 7 k + 6) 6 = ( k + 1)( k + 2)(2 k + 3) 6 = ( k + 1)(( k + 1) + 1)(2( k + 1) + 1) 6 1 We have shown that 1 2 + 2 2 + Â·Â·Â· + ( k + 1) 2 = ( k + 1)(( k + 1) + 1)(2( k + 1) + 1) 6 which means that P ( k + 1) holds. Since P (1) holds and we have shown that P ( k ) implies P ( k + 1), we can conclude by induction that P ( n ) holds for all n . b) 2 n + 3 n is divisible by 5 for each odd n âˆˆ N . Solution: (5 marks). Note that the odd numbers are all numbers of the form 2 k 1 for k âˆˆ N . Let P ( n ) be the statement that 2 2 k 1 + 3 2 k 1 is divisible by 5 for each k âˆˆ N . When n = 1, we have 2 2(1) 1 +3 2(1) 1 = 2+3 = 5 which is clearly divisible by 5. This hhows that P (1) holds. Assume that P ( j ) holds. That is 2 2 j 1 + 3 2 j 1 is divisible by 5. Then 2 2( j +1) 1 + 3 2( j +1) 1 = 2 2 j +1 + 3 2 j +1 = 2 2 Â· 2 2 j 1 + 3 2 Â· 3 2 j 1 = 4 Â· 2 2 j 1 + 9 Â· 3 2 j 1 = [4 Â· 2 2 j 1 + 4 Â· 3 2 j 1 ] + 5 Â· 3 2 j 1 = 4 Â· [2 2 j 1 + 3 2 j 1 ] + 5 Â· 3 2 j 1 The induction hypothesis tells us that 4 Â· [2 2 j 1 +3 2 j 1 ] is divisible by 5. Clearly 5 Â· 3 2 j 1 is divisible by 5. As such P ( j + 1) holds. Using the Principle of Mathematical Induction, we see that P ( k ) holds for all n âˆˆ N . 2) Let a 1 = 1 and for each n â‰¥ 1 let a n +1 = âˆš 3 + 2 a n (This is called a recursively defined sequence.) 2 Prove that for every n âˆˆ N , we have â‰¤ a n â‰¤ a n +1 â‰¤ 3 Solution: (5 marks). Let P ( n ) be the statement that â‰¤ a n â‰¤ a n +1 â‰¤ 3 ....
View
Full
Document
This note was uploaded on 10/21/2010 for the course MATH 147 taught by Professor Wolzcuk during the Fall '09 term at Waterloo.
 Fall '09
 Wolzcuk
 Math

Click to edit the document details