What is the tenth term of this function?
f(1) = 30, f(n) = 3 × f(n − 1) − 90
+26388 What is the tenth term of this function?
f(1) = 30, f(n) = 3 × f(n − 1) − 90
\(\begin{array}{|rcll|} \hline f_n =3f_{n-1} - 90 \qquad f_1 = 30 \qquad f_{10} =\ ? \\ \hline \end{array} \)
\(\small{ \begin{array}{|rcll|} \hline f_1 &=& 30 \\ f_2 &=& 3^1f_1-3^0\cdot 90 \\ f_3 = 3f_2-90 = 3(3f_1-90)-90 &=& 3^2f_1-3^1\cdot90 - 3^0\cdot 90 \\ f_4 = 3f_3-90 = 3(3^2f_1-3\cdot 90 - 90)-90 &=& 3^3f_1-3^2\cdot90- 3^1\cdot90 - 3^0\cdot 90 \\ f_5 = 3f_4-90 = 3(3^3f_1-3^2\cdot 90- 3^1\cdot90 - 3^0\cdot 90)-90 &=& 3^4f_1-3^3\cdot90-3^2\cdot90- 3^1\cdot90 - 3^0\cdot 90 \\ \cdots \\ f_n &=& 3^{n-1}f_1-3^{n-2}\cdot90-3^{n-3}\cdot90-\ldots - 3^0\cdot 90 \\ f_n &=& 3^{n-1}f_1-90\cdot \underbrace{( 3^{n-2} +3^{n-3} +\ldots +1 )}_{=\frac{ 3^{n-1}-1 }{2} }\\ f_n &=& 3^{n-1}f_1-90\cdot \frac{ 3^{n-1}-1 }{2} \\ f_n &=& 3^{n-1}f_1-45\cdot (3^{n-1}-1) \\ \cdots \\ \mathbf{f_n} & \mathbf{=} & \mathbf{-15\cdot(3^{n-1}-3 )} \\ \hline \end{array} }\)
\(\begin{array}{|rcll|} \hline \mathbf{f_n} & \mathbf{=} & \mathbf{-15\cdot(3^{n-1}-3 )} \\\\ f_{10} & =& -15\cdot (3^9-3) \\ f_{10} & =& -15\cdot (19683-3) \\ f_{10} & =& -15\cdot 19680 \\ \mathbf{f_{10}} & \mathbf{=} & \mathbf{ -295200} \\ \hline \end{array}\)
