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.

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

## Comments

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TopNewestThis method is known as "Newton's Sums". – Daniel Liu · 2 years, 9 months ago

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@Aditya Raut What about "Bashing Unbelievable Part 1.5"? – Satvik Golechha · 2 years, 9 months ago

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Any more doubts @Satvik Golechha ??? – Aditya Raut · 2 years, 9 months ago

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-_-– Satvik Golechha · 2 years, 9 months agoLog in to reply

– Abdur Rehman Zahid · 2 years, 2 months ago

It doesn't converge according to Wolfram AlphaLog in to reply

– Aditya Raut · 2 years, 2 months ago

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 !Log in to reply

Similar technics were given in Arthur Engel , i read them little while ago. BTW Nice work Aditya!! – Sanjeet Raria · 2 years, 8 months ago

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Thanks a lot!!! – Aamir Faisal Ansari · 2 years, 9 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, 8 months ago

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Really amazing one... – Sriram Vudayagiri · 2 years, 3 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, 4 months ago

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The

Newton Sumsin 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 · 11 months, 1 week ago

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– Satyabrata Dash · 11 months, 1 week ago

Cool:))Log in to reply

this is cool bro.!! :) – Satyabrata Dash · 11 months, 1 week ago

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