Emps-World Forum
Miscellaneous => Off-Topic => Topic started by: Ameer on March 10, 2017, 03:55:43 pm
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It would be great if you guys can offer some help for these 2 simple math problems that I couldn't find the answer of
Prove that this is right >
1^3 + 2^3 + 3^3 + 4^3 + ........ + n^3 = ( n^2 * (n+1)^2 ) / 4
a Hint, it can be proven to be right with (n+1) ^ 3 to be used as a sum there
more like the final answer should include
The sum to ' What you call it up there ' when it is (n+1)^3 should include that sum I mentioned above + (n+1)^3
(http://imgur.com/BOTuuaY.png)
Second question is like this
Prove that it is right, n times of √2
(http://imgur.com/YgBZOBm.png)
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The answer to this is: Poar neem'n 8)
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Induction can be used on proving the first one.
This is our induction hypothesis: (https://i.gyazo.com/bdbc5575de6d0f9ab6175270dfb050dc.png)
Now we show that it's true for 1 (https://i.gyazo.com/5bf2866403f733149ebc368c787b0658.png)
And now we use the hypothesis to prove it for (n+1)^3
(https://i.gyazo.com/eece83cc793804e1037f5988977c0f57.png)
We get the same result and if it's true for n+1, it's true for n.
I'd say ''Simple'' Math problems is a bit of an understatement. I study maths in university as my secondary subject and we're hardly required to prove any formulas other than a few simple ones with induction. The major students are obviously required to do much more than that.
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@otto, a 'proper' proof by induction format would be
Induction hypothesis: state your formula
Base case: show truth holds for n=1 or whatever your 'first' value in consideration is.
Induction step: "assume the formula holds for some n. we will prove it holds for n+1" and then proceed to show the truth of the n+1 case, as otto did.
customarily, you put a "as a result, by the principle of mathematical induction, the formula holds for all n>=1"
as for your second question, im not quite sure what is being proven. It can be demonstrated that the left hand side, with continuous repetition of the insides, does in fact equal 2.
If you're trying to show that the nth repetition is less than 2, then you use induction, similar to above.
the base case obviously holds : sqrt(2) < 2
assume sqrt(2+.... (n times) ) < 2
then sqrt(2 + ... (n+1 times)) = sqrt(2+sqrt(2+ .... (n times))) < sqrt(2+2)=2
Hence the results holds for all finite, positive n.