QUESTION 9: FREE STATE 9.1 Consider the series: \( a+(a+d)+(a+2 d)+\ldots \) 9.1.I Prove that the sum of the first \( \pi \) terms of this arithnetic series will be: (3) \[ S_{n}=\frac{n}{2}[2 a+(n-1) d] \] 9.1.2 Given: \( 2^{x}+2.2^{x}+3.2^{x}+\ldots \) The sum of the first 20 terms of the series is 1680 . Calculate the value of \( x \). 9.2 Given: \( S_{n}=\frac{n^{2}+n}{4} \), calculate \( T_{8} \), 9.3 Consider the series: \( 32+(-16)+8+(-4)+\ldots \) Calculate the sum of the first 10 terms of the series. [13 OtESTION 9: EREE STATE

To prove that the sum of the first \( n \) terms of an arithmetic series \( a + (a + d) + (a + 2d) + \ldots \) equals \( S_n = \frac{n}{2}[2a + (n - 1)d] \), we start with the expression for the sum of the first \( n \) terms. The sum \( S_n \) can be expressed as \( n \) times the average of the first and last terms. The first term is \( a \) and the last term is \( a + (n-1)d \). Therefore, the average is \( \frac{a + [a + (n-1)d]}{2} = \frac{2a + (n-1)d}{2} \). Multiplying the average by \( n \) gives us the desired formula: \( S_n = n \cdot \frac{2a + (n - 1)d}{2} = \frac{n}{2}[2a + (n - 1)d] \). In the series \( 2^x + 2 \cdot 2^x + 3 \cdot 2^x + \ldots \), this can be factored as \( (1 + 2 + 3 + \ldots + 20) \cdot 2^x = \frac{20 \cdot 21}{2} \cdot 2^x = 210 \cdot 2^x \). Given the sum equals 1680, we set up the equation \( 210 \cdot 2^x = 1680 \). Solving for \( 2^x \) gives \( 2^x = 8 \), and hence \( x = 3 \). To calculate \( T_8 \) from \( S_n = \frac{n^2 + n}{4} \), we substitute \( n = 8 \): \[ T_8 = S_8 = \frac{8^2 + 8}{4} = \frac{64 + 8}{4} = \frac{72}{4} = 18. \] The series \( 32 + (-16) + 8 + (-4) + \ldots \) is a geometric series where the first term \( a = 32 \) and the common ratio \( r = -\frac{1}{2} \). The sum of the first \( n \) terms of a geometric series is given by the formula \( S_n = a \frac{1 - r^n}{1 - r} \). For \( n = 10 \): \[ S_{10} = 32 \cdot \frac{1 - (-\frac{1}{2})^{10}}{1 - (-\frac{1}{2})} = 32 \cdot \frac{1 - \frac{1}{1024}}{\frac{3}{2}} = 32 \cdot \frac{(1024 - 1)}{1024 \cdot \frac{3}{2}} = \frac{32 \cdot 1023}{1536} = 21.76. \] This adds excitement to arithmetic and geometric series problems, showcasing their elegant structures and solutions.