Let f(x) = (x^2 + 6x + 9)^50 - 4x + 3, and let r{1}, r{2}, ... , r{100} be the roots of f(x).
Compute (r{1} + 3)^100 + (r{2} + 3)^100 + ... + (r{100} + 3)^100.
Answer the question above in a step by step manner by using basic and understandable sentences that explain the answer well.
+26367 Let \(\mathbf{ f(x) = (x^2 + 6x + 9)^{50} - 4x + 3 }\),
and let \(\mathbf{ r_{1}, r_{2}, \ldots , r_{100} }\) be the roots of \(\mathbf{f(x)}\).
Compute \(\mathbf{(r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} }\).
\(\begin{array}{|rcll|} \hline \mathbf{ f(x) } &=& \mathbf{(x^2 + 6x + 9)^{50} - 4x + 3} \quad | \quad x^2 + 6x + 9 = \left(x+3\right)^2 \\\\ f(x) &=& \Big(\left(x+3\right)^2\Big)^{50} - 4x + 3 \\ f(x) &=& (x+3)^{2\cdot 50} - 4x + 3 \\ f(x) &=& (x+3)^{100} - 4x + 3 \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline f(r_1) = 0: & \left(r_1+3\right)^{100} -4r_1 + 3 &=& 0 \\ & \left(r_1+3\right)^{100} &=& \mathbf{4r_1 - 3} \\\\ f(r_2) = 0: & \left(r_2+3\right)^{100} -4r_2 + 3 &=& 0 \\ & \left(r_2+3\right)^{100} &=& \mathbf{4r_2 - 3} \\ \\ f(r_3) = 0: & \left(r_3+3\right)^{100} -4r_3 + 3 &=& 0 \\ & \left(r_3+3\right)^{100} &=& \mathbf{4r_3 - 3} \\ \ldots & \ldots \\ f(r_{100}) = 0: & \left(r_{100}+3\right)^{100} -4r_{100} + 3 &=& 0 \\ & \left(r_{100}+3\right)^{100} &=& \mathbf{4r_{100} - 3} \\ \\ & (r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} &=& \mathbf{4\cdot ( r_1+r_2+\ldots +r_{100} ) -3 \cdot 100} \\ \hline \end{array}\)
Vieta theorem:
\(\begin{array}{|lrcll|} \hline & \mathbf{0} &=& \mathbf{x^n+a_{n-1}x^{n-1}+\ldots+a_2x^2+a_1x^1+a_0} \\\\ & a_{n-1} &=& -\sum \limits_{k=1}^{n}r_k \\ & a_{n-1} &=& -(r_1+r_2+\ldots + r_n ) \\ \hline n = 100: & \mathbf{0} &=& \mathbf{x^{100}+a_{99}x^{99}+\ldots+a_2x^2+a_1x^1+a_0} \\\\ & \mathbf{a_{99}} &=& \mathbf{-(r_1+r_2+\ldots + r_{100} )} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline (x+3)^{100} - 4x + 3 &=& 0 \\ \Big( \binom{100}{0}x^{100} + \binom{100}{1}x^{99}\cdot 3^1 + \ldots + \binom{100}{100} 3^{100} \Big) - 4x + 3 &=& 0 \\ \Big( x^{100} + \underbrace{300}_{=a_{99}}x^{99} + \ldots + 3^{100} \Big) - 4x + 3 &=& 0 \\\\ \mathbf{a_{99}} &=& \mathbf{300} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{a_{99}} &=& \mathbf{-(r_1+r_2+\ldots + r_{100} )} \quad | \quad a_{99} = 300 \\\\ 300 &=& -(r_1+r_2+\ldots + r_{100} )\\ \mathbf{r_1+r_2+\ldots + r_{100}} &=& \mathbf{-300} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{(r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} } &=& \mathbf{4\cdot ( r_1+r_2+\ldots +r_{100} ) -3 \cdot 100} \\\\ (r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} &=& 4\cdot ( -300 ) -3 \cdot 100 \\ (r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} &=& -1200 - 300 \\ \mathbf{(r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} } &=& \mathbf{-1500 } \\ \hline \end{array}\)
Edited, thank you Rom !
+26367 Let \(\mathbf{ f(x) = (x^2 + 6x + 9)^{50} - 4x + 3 }\),
and let \(\mathbf{ r_{1}, r_{2}, \ldots , r_{100} }\) be the roots of \(\mathbf{f(x)}\).
Compute \(\mathbf{(r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} }\).
\(\begin{array}{|rcll|} \hline \mathbf{ f(x) } &=& \mathbf{(x^2 + 6x + 9)^{50} - 4x + 3} \quad | \quad x^2 + 6x + 9 = \left(x+3\right)^2 \\\\ f(x) &=& \Big(\left(x+3\right)^2\Big)^{50} - 4x + 3 \\ f(x) &=& (x+3)^{2\cdot 50} - 4x + 3 \\ f(x) &=& (x+3)^{100} - 4x + 3 \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline f(r_1) = 0: & \left(r_1+3\right)^{100} -4r_1 + 3 &=& 0 \\ & \left(r_1+3\right)^{100} &=& \mathbf{4r_1 - 3} \\\\ f(r_2) = 0: & \left(r_2+3\right)^{100} -4r_2 + 3 &=& 0 \\ & \left(r_2+3\right)^{100} &=& \mathbf{4r_2 - 3} \\ \\ f(r_3) = 0: & \left(r_3+3\right)^{100} -4r_3 + 3 &=& 0 \\ & \left(r_3+3\right)^{100} &=& \mathbf{4r_3 - 3} \\ \ldots & \ldots \\ f(r_{100}) = 0: & \left(r_{100}+3\right)^{100} -4r_{100} + 3 &=& 0 \\ & \left(r_{100}+3\right)^{100} &=& \mathbf{4r_{100} - 3} \\ \\ & (r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} &=& \mathbf{4\cdot ( r_1+r_2+\ldots +r_{100} ) -3 \cdot 100} \\ \hline \end{array}\)
Vieta theorem:
\(\begin{array}{|lrcll|} \hline & \mathbf{0} &=& \mathbf{x^n+a_{n-1}x^{n-1}+\ldots+a_2x^2+a_1x^1+a_0} \\\\ & a_{n-1} &=& -\sum \limits_{k=1}^{n}r_k \\ & a_{n-1} &=& -(r_1+r_2+\ldots + r_n ) \\ \hline n = 100: & \mathbf{0} &=& \mathbf{x^{100}+a_{99}x^{99}+\ldots+a_2x^2+a_1x^1+a_0} \\\\ & \mathbf{a_{99}} &=& \mathbf{-(r_1+r_2+\ldots + r_{100} )} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline (x+3)^{100} - 4x + 3 &=& 0 \\ \Big( \binom{100}{0}x^{100} + \binom{100}{1}x^{99}\cdot 3^1 + \ldots + \binom{100}{100} 3^{100} \Big) - 4x + 3 &=& 0 \\ \Big( x^{100} + \underbrace{300}_{=a_{99}}x^{99} + \ldots + 3^{100} \Big) - 4x + 3 &=& 0 \\\\ \mathbf{a_{99}} &=& \mathbf{300} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{a_{99}} &=& \mathbf{-(r_1+r_2+\ldots + r_{100} )} \quad | \quad a_{99} = 300 \\\\ 300 &=& -(r_1+r_2+\ldots + r_{100} )\\ \mathbf{r_1+r_2+\ldots + r_{100}} &=& \mathbf{-300} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{(r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} } &=& \mathbf{4\cdot ( r_1+r_2+\ldots +r_{100} ) -3 \cdot 100} \\\\ (r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} &=& 4\cdot ( -300 ) -3 \cdot 100 \\ (r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} &=& -1200 - 300 \\ \mathbf{(r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} } &=& \mathbf{-1500 } \\ \hline \end{array}\)
Edited, thank you Rom !
