Newton is tired now!

Algebra Level 5

If $a + b + c + d + e = 1$

$a^2 + b^2 + c^2 + d^2 + e^2 = 1$

$a^3 + b^3 + c^3 + d^3 + e^3 = 2$

$a^4 + b^4 + c^4 + d^4 + e^4 = 3$

$a^5 + b^5 + c^5 + d^5 + e^5 = 5$

then find the value of $30(a^7 + b^7 + c^7 + d^7 + e^7)$

This problem is inspired by many problems

The answer is 233.

1 solution

Chew-Seong Cheong
Nov 4, 2015

$\begin{aligned} \text{Let } P_n & = a^n+b^n+c^n+d^n+e^n, \text{ where } n = 1,2,3... \\ S_1 & = a+b+c+d+e = P_1 \\ S_2 & = ab+ac+ad+ae+bc++bd+be+cd+ce+de \\ S_3 & = abc+abd+abe+acd+ace+ade+bcd+ bce+bde+cde \\ S_4 & = abcd+abde+acde+bcde \\ S_5 & = abcde \end{aligned}$

Using Newton sums method, we have:

$P_k = \begin{cases} S_1P_{k-1}-S_2P_{k-2}+S_3P_{k-3}-...(-1)^{k-1}kS_k & \text{for } k \le 5 \\ S_1P_{k-1}-S_2P_{k-2}+S_3P_{k-3}-S_4P_{k-4}+S_5P_{k-5} & \text{for } k > 5 \end{cases}$

Given $P_1 = 1, P_2 = 1, P_3 = 2, P_4 = 3, P_5 = 5$ , we find that $S_1 = 1, S_2 = 0, S_3 = \frac{1}{3}, S_4 = - \frac{1}{6}, S_5 = \frac{3}{10}$ .

Using an Excel spreadsheet (see below) for ease of computation, we find that $P_6 = \frac{92}{15}$ and $P_7 = a^7+b^7+c^7+d^7+e^7 = \frac{233}{30}\quad \Rightarrow 30(a^7+b^7+c^7+d^7+e^7) = \boxed{233}$

Did the same way!!😀😀

Anurag Pandey - 4 years, 10 months ago

Perfect Solution!!

Dev Sharma - 5 years, 7 months ago

Given the title and the fibonnaci series, I was expecting something other than Newton's Sums.

shaurya gupta - 5 years, 7 months ago

Same, is there a differentt way of solving

Mardokay Mosazghi - 5 years, 6 months ago

@Mardokay Mosazghi – Of course, with 5 unknown and 5 equations, you can find out the values of a, b, c, d, e and f, but that will be difficult.

Chew-Seong Cheong - 5 years, 6 months ago

Newton is tired now!

The answer is 233.

1 solution

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