(a) Prove by MI that n^3+(n+1)^3+(n+2)^3 is divisible by 9 for any positive integers n. [Only have to do when n=k+1 if P(k) is assumed to be true for n=k.]
(b) The lengths of three cubes are three consecutive even numbers and the length of the smallest cube is 2a, where a is an integer. Using the result in (a), show that the total volume of the three cubes is a multipe of 72.
Induction
2010-04-17 11:19 pm
回答 (1)
2010-04-17 11:41 pm
✔ 最佳答案
a)
Let P(k) is true for some positive integers k.
k³+(k+1)³+(k+2)³=9M ,where M is an integer.
For n=k+1:
(k+1)³+(k+2)³+(k+3)³
=[9M−k³]+(k+3)³
=[9M−k³]+(k³+9k²+27k+27)
=9M+9k²+27k+27 ,which is divisible by 9.
b)
Total volume of the three cubes
=(2a)³+(2a+2)³+(2a+4)³
=8a³+8(a+1)³+8(a+2)³
=8[a³+(a+1)³+(a+2)³]
=8(9N) ,where N is an integer.
=72N ,which is divisible by 72.
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