Inspired By Someone!

Algebra Level 4

Consider the following equations $\begin{aligned} &a_{1}+a_{2}+a_{3} &=& 6 \\ &a_{1}^2+a_{2}^2+a_{3}^2&=&194 \\ &a_{1}^3+a_{2}^3+a_{3}^3&=&126 \\ &a_{1}^4+a_{2}^4+a_{3}^4&=&13058 \\ \end{aligned}$

Find $a_{1}^5+a_{2}^5+a_{3}^5$ .

The answer is -9474.

1 solution

Alan Enrique Ontiveros Salazar
Dec 10, 2014

Let $S_1=a_1+a_2+a_3$ , $S_2=a_1a_2+a_1a_3+a_2a_3$ , $S_3=a_1a_2a_3$ and $P_n=a_1^n+a_2^n+a_3^n$ . By Newton's sums we know:

$P_1=S_1$

$P_2=S_1^2-2S_2$

$P_3=S_1^3-3S_1S_2+3S_3$

$P_n=S_1P_{n-1}-S_2P_{n-2}+S_3P_{n-3}$

With the given values the following system is formed:

$S_1=6$

$S_1^2-2S_2=194$

$S_1^3-3S_1S_2+3S_3=126$

Solving that we get $S_1=6$ , $S_2=-79$ and $S_3=-504$ .

So:

$P_n=6P_{n-1}+79P_{n-2}-504P_{n-3}$

$P_4=6P_3+79P_2-504P_1=6(126)+79(194)-504(6)=13058$

$P_5=6P_4+79P_3-504P_2=6(13058)+79(126)-504(194)=\boxed{-9474}$

Excellent :)

Parth Lohomi - 6 years, 6 months ago

@Parth Lohomi there is no need to mention $a_{1}^{4}+a_{2}^{4}+a_{3}^{4} =13058$

Shubhendra Singh - 6 years, 6 months ago

I also agree to you.

Anshuman Singh Bais - 5 years, 9 months ago

@Parth Lohomi change the answer to 9473!!

Adarsh Kumar - 6 years, 6 months ago

Given that phrasing of the problem was unclear, those who previously answered 9473 have been marked correct.

I have updated the phrasing of the problem and the answer is now -9474.

Calvin Lin Staff - 6 years, 6 months ago

Inspired By Someone!

The answer is -9474.

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