+118723 If f(1)=3 and f(n)=-2f(n-1)+1, then f(5) is equal to?
f(2)=-2*f(1)+1=-2*3+1=-5
f(3)=-2*f(2)+1=-2*-5+1=-2*-5+1=11
f(4)=-2*f(3)+1=-2*11+1=-21
f(5)=-2*f(4)+1=-2*-21+1=43
+26400 If f(1)=3 and f(n)=-2f(n-1)+1, then f(5) is equal to?
\(\small{ \begin{array}{rclcl} f(1)&& &=&3\\ f(2)&& &=& (-2)^1\cdot f(1)+1 \\ f(3)&=& (-2)\cdot [ f(2)]+1 \\ &=& (-2)\cdot[ (-2)\cdot f(1)+1 ]+1 &=& (-2)^2\cdot f(1) + (-2)^1 + 1\\ f(4)&=& (-2)\cdot [ f(3)]+1 \\ &=& (-2)\cdot[ (-2)^2\cdot f(1) + (-2) + 1 ]+1 &=& (-2)^3\cdot f(1) + (-2)^2 + (-2)^1 + 1\\ f(5)&=& (-2)\cdot [ f(4)]+1 \\ &=& (-2)\cdot[ (-2)^3\cdot f(1) + (-2)^2 + (-2) + 1 ]+1 &=& (-2)^4\cdot f(1) + (-2)^3 + (-2)^2 + (-2)^1 + 1\\ f(5)&& &=&16\cdot 3 -8 + 4 -2 + 1\\ f(5)&& &=&48 -8 + 4 -2 + 1\\ f(5)&& &=&43\\ \end{array} } \)
\(\begin{array}{rclcl} f(n)&=& (-2)^{n-1}\cdot f(1) + (-2)^{n-2} + (-2)^{n-3} + (-2)^{n-4}+\dots + (-2)^1 + 1\\\\ && \text{sum of a geometric sequence } ~r=-2 \quad a_1 = 1\\ && (-2)^{n-2} + (-2)^{n-3} + (-2)^{n-4}+\dots + (-2)^1 + 1 = \frac{1-(-2)^{n-1}}{1-(-2)}\\\\ f(n)&=& (-2)^{n-1}\cdot f(1) + \frac{1-(-2)^{n-1}}{3}\\ f(n)&=& (-2)^{n-1}\cdot f(1) + \frac13 - \frac{(-2)^{n-1}}{3}\\ f(n)&=& (-2)^{n-1}\cdot \left[ f(1) -\frac13 \right] + \frac13 \\ \end{array} \)
\(\boxed{ \begin{array}{rcll} f(n)&=& (-2)^{n-1}\cdot \left[ f(1) -\frac13 \right] + \frac13 \end{array} } \)
\(\begin{array}{rcll} f(5)&=& (-2)^{5-1}\cdot \left[ f(1) -\frac13 \right] + \frac13 \qquad | \qquad f(1) = 3 \\ f(5)&=& (-2)^{4}\cdot \left[ 3 -\frac13 \right] + \frac13 \\ f(5)&=& 16\cdot \left[ 3 -\frac13 \right] + \frac13 \\ f(5)&=& 16\cdot \left[ \frac83 \right] + \frac13 \\ f(5)&=& \frac{16\cdot 8 + 1}{3} \\ f(5)&=& \frac{129}{3} \\ \mathbf{f(5)}&\mathbf{=}& \mathbf{43} \end{array} \)
