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An awesome generalisation- of Bashing Unavailable

If you were given the three equations \[\color{Blue}{a+b+c=1}\]\[\color{Blue}{a^2+b^2+c^2=2}\] \[\color{Blue}{a^3+b^3+c^3=3}\]

And then the values of \(\color{Red}{a^n+b^n+c^n}\) are asked for some \(n \in \mathbb{Z}^+\), then you can get the answer by long expansions, squaring, getting needed terms etc, but there is a smart way too !

Some days ago, i had shared a set by name Bashing Unavailable...Awesome problems , which was on problems of this type , for \(n \in \{ 4,5,6,7,8,9,10\}\).

If you haven't tried the set, I prefer telling you to try it before you read this note.


You can get the thing for \(n=4\) by algebraic manipulations (and also for further values of \(n\) ) , for example this "part of the solution" which I knowingly wrote to the first problem of the set.

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This way, you can get the values.... but here goes the smarter way, the \(\color{Purple}{\textbf{generalisation}}\). My way is similar to my friend Aamir Faisal Ansari,a worth following person's way...


See that as shown in the image, we can get values of \(ab+bc+ca=\dfrac{-1}{2}\) and \(abc=\dfrac{1}{6}\)


Next, let's define a sequence \(\{t_n\}\) as \(t_n=a^n+b^n+c^n\)

Then, we do the following algebraic manipulation which will give us the generalisation

\(a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1}) - \text{<something>}\)

(something = the extra terms that will come in the expansion of first term of the RHS)

\(a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1}) \\ \quad \quad \quad \quad -(ac^{n-1} + bc^{n-1}+ab^{n-1}+cb^{n-1}+ba^{n-1}+ca^{n-1})\)

For the extra terms, we have

\((ac^{n-1} + bc^{n-1}+ab^{n-1}+cb^{n-1}+ba^{n-1}+ca^{n-1}) = \\ \quad \quad \quad \quad \quad \quad (ab+bc+ca)(a^{n-2}+b^{n-2}+c^{n-2}) - \text{<other something>}\)

\(\textbf{Other something}= bca^{n-2}+cab^{n-2}+abc^{n-2} = abc(a^{n-3}+b^{n-3}+c^{n-3})\)

Thus, we have

\(a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})\\\quad \quad \quad\quad \quad - (ab+bc+ca)(a^{n-2}+b^{n-2}+c^{n-2}) \\ \quad \quad \quad \quad\quad+ abc(a^{n-3}+b^{n-3}+c^{n-3})\)

From this we get the recurrence relation \[t_n=(a+b+c) t_{n-1} -(ab+bc+ca)t_{n-2} +(abc)t_{n-3}\]

And because we know the values of all the coefficients in this thing, we get the recurrence relation \[t_n=t_{n-1} +\frac{1}{2} t_{n-2}+\frac{1}{6} t_{n-3}\]

From this recurrence relation, \(\color{Green}{\text{because you already know the first 3 terms}}\), you can get the value of all the further terms like a cakewalk !

Reshare if this is helpful, and try to solve the last part of it, the \(6^{th}\) part of the set - Bashing Unavailable Part 6

Note by Aditya Raut
2 years, 11 months ago

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This method is known as "Newton's Sums". Daniel Liu · 2 years, 11 months ago

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@Aditya Raut What about "Bashing Unbelievable Part 1.5"? Satvik Golechha · 2 years, 11 months ago

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@Satvik Golechha It will come after I release \(\displaystyle \color{Purple}{\textbf{Bashing Unavailable part }} \color{Green}{\sqrt{\dfrac{\sum_{k=\sqrt{2}} ^{\sqrt{17}} \bigl( 2^{\sqrt{k}} + k^{\sqrt{2}} \bigr)}{\int_0 ^\infty \sin^{2014} x \text{ dx} - \log_{9.7} 888 }}}\)

Any more doubts @Satvik Golechha ??? Aditya Raut · 2 years, 11 months ago

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@Aditya Raut -_- Satvik Golechha · 2 years, 11 months ago

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@Aditya Raut It doesn't converge according to Wolfram Alpha Abdur Rehman Zahid · 2 years, 4 months ago

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@Abdur Rehman Zahid Oh man! That was just a troll, because I was meaning to tell Satvik there's no 1.5 coming up, this thing was just for integers... Anyway, thanks for telling, but that big term is not seriously typed ! Aditya Raut · 2 years, 4 months ago

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Similar technics were given in Arthur Engel , i read them little while ago. BTW Nice work Aditya!! Sanjeet Raria · 2 years, 10 months ago

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Thanks a lot!!! Aamir Faisal Ansari · 2 years, 11 months ago

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Thanks a lott!! @Aditya Raut !! A wonderful Note!! i have been struggling at questions of these type a bit...but not anymore...thanks!! :):) Abhineet Nayyar · 2 years, 10 months ago

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Really amazing one... Sriram Vudayagiri · 2 years, 5 months ago

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What's amazing about the set is that even though you have so many equations in the same three variables, you can't find their values. I mean, I know you're transforming the given equations to the new ones, so you technically can't find the values of the variables, but still. Omkar Kulkarni · 2 years, 5 months ago

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The Newton Sums in 3 variables are:

\[p=a+b+c \\ q=ab+bc+ac \\ r=abc\]

These are fairly straightforward to calculate if you're give \(\sum a,\sum a^2,\sum a^3\).

If you let \(a,b,c\) be the roots of the monic polynomial \(f(x)\) then you have:

\[f(x)=(x-a)(x-b)(x-c)=x^3-px^2+qx-r \quad \quad f(a)=f(b)=f(c)=0 \\ f(a)=0 \implies a^3-pa^2+qa-r=0 \implies a^3=pa^2-qa+r\]

If we let \(t_n=\sum a^n\) then we get:

\[a^3=pa^2-qa+r \implies a^{n+3}=p a^{n+2}-q a^{n+1}+r a^n \\ \sum a^{n+3}=p \sum a^{n+2} -q \sum a^{n+1}+r \sum a^n \\ t_{n+3}=p \; t_{n+2}-q \; t_{n+1}+r \; t_n\]

This gives you a way of quickly calculating \(t_n\) for large \(n\) as you will know \(p,q,r\).

I've used this method on this problem. Sam Bealing · 1 year, 1 month ago

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@Sam Bealing Cool:)) Satyabrata Dash · 1 year, 1 month ago

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this is cool bro.!! :) Satyabrata Dash · 1 year, 1 month ago

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