Hi,
I suggest writing and then using the binomial theorem to expand .
To enter math. equations you can use LaTeX. (see http://www.mathhelpforum.com/math-help/latex-help/)
hi folks,
I hope this is the right place to post this! I have no idea how to proceeed with the following question:
Show that for all positive integral values of n
[HTML]7 <sup>n</sup> + 2 <sup>2n+1</sup>[/HTML]
is divisible by 3.
I tried a few terms as follows:
n = 1. 7 ^ 1 + 2 ^ 3 = 7 + 8 = 15
n = 2. 7 ^ 2 + 2 ^ 5 = 49 + 32 = 81
n = 3. 7 ^ 3 + 2 ^ 7 = 343 + 128 = 471
and these are all divisible by 3 but how do I handle the general case?
I thought of expanding the expression i.e.
2.2 ^ n. 2 ^ n - (1 - 8) ^ n
and using a binomial on the second term but it doesn't get me anywhere. I guess I am trying to calculate the sum of a series and show that it has a factor of 3 but I can't see how to do it. Any ideas?
regards and thanks
Simon
p.s. sorry about the formating. the HTML stuff doesn't seem to come out so I resorted to the ^ symbol which is pretty unpretty!
Hi,
I suggest writing and then using the binomial theorem to expand .
To enter math. equations you can use LaTeX. (see http://www.mathhelpforum.com/math-help/latex-help/)
Hello, s_ingram!
How about an inductive proof?
Show that for any positive integer is divisible by 3.
Verify ... divisible by 3.
Assume
Add to both sides:
. .
. . . a multiple of 3
. . . . . . . .
Therefore: . .
We have proved
. . The inductive proof is complete.
Hi Soroban, now that is a smart proof. Too smart! How did you think adding those two terms? Once you see them it all fits but finding them is the real trick! I have never really been impressed with proofs by induction, but I am now! Usually I just try to add the next term in the series and ensure that it has a corresponding impact on the expression fo the sum but with 7 sup k+1 + 2 sup 2k+1 I couldn't even imagine what the next term would be!