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January 31, 2015

January 31, 2015

Posted by **Hjm** on Saturday, August 17, 2013 at 9:16pm.

- Math -
**Reiny**, Saturday, August 17, 2013 at 10:24pmstep 1:

test for n = 1

1^3 + 2(1) = 3 , which is a multiple of 3

step 2:

assume it is true for n = k

that is, k^3 + 3k is a multiple of 3, or it is divisible by 3

step 3:

show that it is also true for n = k+1

that is, show that (k+1)^3 + 3k is a multiple of 3

let's take the difference

(k+1)^3 + 2(k+1) - (k^3 + 2k)

= k^3 + 3k^2 + 3k + 1 + 2k + 2 - k^3 - 2k

= 3k^2 + 3k + 3

= 3(k^2 + k + 1)

which is divisible by 3, (since 3 is a factor)

so n^3 + 3n is always a multiple of 3

the property I used is the following:

if 2 numbers are divisible by the same number, then their difference is divisible by that same number

e.g. 91 and 49 are both divisible by 7

then 91-49 or 42 is also divisible by 7

-- try it for other numbers.

since we knew the second number, k^3 + 2k , was divisible by 3 and the result was divisible by 3, then the first number, (k+1)^3 + 2(k+1) has to be divisible by 3

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