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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThis method is known as "Newton's Sums".

Log in to reply

@Aditya Raut What about "Bashing Unbelievable Part 1.5"?

Log in to reply

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 ???

Log in to reply

-_-Log in to reply

It doesn't converge according to Wolfram Alpha

Log in to reply

Log in to reply

Thanks a lot!!!

Log in to reply

Similar technics were given in Arthur Engel , i read them little while ago. BTW Nice work Aditya!!

Log in to reply

Thanks a lott!! @Aditya Raut !! A wonderful Note!! i have been struggling at questions of these type a bit...but not anymore...thanks!! :):)

Log in to reply

Really amazing one...

Log in to reply

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.

Log in to reply

this is cool bro.!! :)

Log in to reply

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.

Log in to reply

Cool:))

Log in to reply