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# Let , and let be the roots of . Compute ​.

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Let $$f(x)=(x^2+6x+9)^{50}-4x+3$$, and let $$r_1,r_2,\ldots,r_{100}$$ be the roots of $$f(x)$$.

Compute $$(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}$$.

May 13, 2019

#1
+23878
+2

Let
$$f(x)=(x^2+6x+9)^{50}-4x+3$$, and let $$r_1,r_2,\ldots,r_{100}$$ be the roots of $$f(x)$$.
Compute $$(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}$$.

$$\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x^2+6x+9)^{50}-4x+3} \\ &=& \left((x+3)^2\right)^{50}-4x+3 \\ \mathbf{f(x)} &=& \mathbf{\left( x+3 \right)^{100}-4x+3} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & f(r_1) =0 &=& \left( r_1+3 \right)^{100}-4r_1+3 \\ & f(r_2) =0 &=& \left( r_2+3 \right)^{100}-4r_2+3 \\ & \ldots \\ & f(r_{100}) =0 &=& \left( r_{100}+3 \right)^{100}-4r_{100}+3 \\ & \hline \\ \text{sum} & 0 &=& (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} \\ & && -4(r_1+r_2+\ldots + r_{100}) + 3\cdot 100 \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ \hline \end{array}$$

$$\mathbf{\text{vieta:}}$$

For any polynomial equation
$$0=x^n+a_{n-1}·x^{n-1}+...+a_2·x^2+a_1·x^1+a_0$$
with the solutions $$r_1\dots r_n$$, the relatively simple formulas for $$a_0$$ and $$a_{n-1}$$ are:

$$a_0=(-1)^n \prod\limits_{k=1}^{n} r_k \\ a_{n-1}= -\sum \limits_{k=1}^{n} r_k$$

$$\begin{array}{|rcll|} \hline && \left( x+3 \right)^{100} \\ \\ &=& x^{100} + \binom{100}{1}x^{99}\cdot 3 + \ldots \\ \\ &=& x^{100} + \underbrace{300}_{ \underbrace{=a_{n-1}}_{= -(r_1+r_2+\ldots + r_{100})}} x^{99} + \ldots \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ && \boxed{r_1+r_2+\ldots + r_{100} = -300} \\ (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(-300) - 300 \\ \mathbf{(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}} &=& \mathbf{-1500} \\ \hline \end{array}$$

May 14, 2019
edited by heureka  May 14, 2019

#1
+23878
+2

Let
$$f(x)=(x^2+6x+9)^{50}-4x+3$$, and let $$r_1,r_2,\ldots,r_{100}$$ be the roots of $$f(x)$$.
Compute $$(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}$$.

$$\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x^2+6x+9)^{50}-4x+3} \\ &=& \left((x+3)^2\right)^{50}-4x+3 \\ \mathbf{f(x)} &=& \mathbf{\left( x+3 \right)^{100}-4x+3} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & f(r_1) =0 &=& \left( r_1+3 \right)^{100}-4r_1+3 \\ & f(r_2) =0 &=& \left( r_2+3 \right)^{100}-4r_2+3 \\ & \ldots \\ & f(r_{100}) =0 &=& \left( r_{100}+3 \right)^{100}-4r_{100}+3 \\ & \hline \\ \text{sum} & 0 &=& (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} \\ & && -4(r_1+r_2+\ldots + r_{100}) + 3\cdot 100 \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ \hline \end{array}$$

$$\mathbf{\text{vieta:}}$$

For any polynomial equation
$$0=x^n+a_{n-1}·x^{n-1}+...+a_2·x^2+a_1·x^1+a_0$$
with the solutions $$r_1\dots r_n$$, the relatively simple formulas for $$a_0$$ and $$a_{n-1}$$ are:

$$a_0=(-1)^n \prod\limits_{k=1}^{n} r_k \\ a_{n-1}= -\sum \limits_{k=1}^{n} r_k$$

$$\begin{array}{|rcll|} \hline && \left( x+3 \right)^{100} \\ \\ &=& x^{100} + \binom{100}{1}x^{99}\cdot 3 + \ldots \\ \\ &=& x^{100} + \underbrace{300}_{ \underbrace{=a_{n-1}}_{= -(r_1+r_2+\ldots + r_{100})}} x^{99} + \ldots \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ && \boxed{r_1+r_2+\ldots + r_{100} = -300} \\ (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(-300) - 300 \\ \mathbf{(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}} &=& \mathbf{-1500} \\ \hline \end{array}$$

heureka May 14, 2019
edited by heureka  May 14, 2019
#2
+133
0

Thank you i understand but for Vieta's I learned that (-1)^k=(an-k/an) is this the same or sound familiar

May 16, 2019