$a^n+b^n+c^n+d^n$

Algebra Level 3

$\begin{aligned} a\ +\ b\ +\ c\ +\ d\ & = 1 \\ a^2+b^2+c^2+d^2 & = 2 \\ a^3+b^3+c^3+d^3 & = 3 \\ a^4+b^4+c^4+d^4 & = 4 \\ a^5+b^5+c^5+d^5 & = p \end{aligned}$

Find $24p$ .

The answer is 139.

2 solutions

Chew-Seong Cheong
Mar 17, 2020

Let $P_n = a^n+b^n+c^n+d^n$ , where $n$ is a positive integer, and $S_1 = a+b+c+d$ , $S_2 = ab+ac+ad+bc+bd+cd$ , $S_3=abc+abd+acd+bcd$ , and $S_4=abcd$ . Using Newton's sums (identities) , we have:

$\begin{aligned} P_1 & = S_1 = 1 \\ P_2 & = S_1P_1 - 2S_2 = 1-2S_1 = 2 & \small \blue{\implies S_2 = - \frac 12} \\ P_3 & = S_1P_2 - S_2P_1+3S_3 = 2+\frac 12 + 3S_3 = 3 & \small \blue{\implies S_3 = \frac 16} \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 -4S_4 = 3+1 + \frac 16 -4S_4 = 4 & \small \blue{\implies S_4 = \frac 1{24}} \\ P_5 & = S_1P_4 - S_2P_3 + S_3P_2 -S_4P_1 = 4 + \frac 32 + \frac 13 - \frac 1{24} = \frac {139}{24} \end{aligned}$

Therefore, $24p=24 \times \dfrac {139}{24} = \boxed{139}$ .

Vaibhav Priyadarshi
Mar 16, 2020

$a^n+b^n+c^n+d^n$

The answer is 139.

2 solutions

1 pending report

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a n + b n + c n + d n a^n+b^n+c^n+d^n a n + b n + c n + d n

The answer is 139.

2 solutions

1 pending report

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$a^n+b^n+c^n+d^n$