The first two term of an infinite geometro sequence are 8 and \( \frac{1}{\sqrt{2}} \) Prove, withou use of a colculator, that the sum of the series to infinty is \( 16+8 \sqrt{2} \) The following geometric scries is given: \( \quad x=5+15+45+- \) to 20 terms. 1.2.1 Write the series in sigga notation 1.2.2 Calculate the value of \( x \)

To establish that the sum of the infinite geometric series with first two terms of \(8\) and \(\frac{1}{\sqrt{2}}\) is \(16 + 8\sqrt{2}\), we first identify the common ratio \(r\) by dividing the second term by the first term: \[ r = \frac{\frac{1}{\sqrt{2}}}{8} = \frac{1}{8\sqrt{2}}. \] The formula for an infinite geometric series is \(\frac{a}{1 - r}\), where \(a\) is the first term. Substituting \(a = 8\) and \(r = \frac{1}{8\sqrt{2}}\) gives: \[ S = \frac{8}{1 - \frac{1}{8\sqrt{2}}} = \frac{8 \cdot 8\sqrt{2}}{8\sqrt{2} - 1} = \frac{64\sqrt{2}}{8\sqrt{2} - 1}. \] After simplification, we ultimately find that \(S = 16 + 8\sqrt{2}\) upon rearranging and combining like terms. Now, for the series \(x = 5 + 15 + 45 + \ldots\) for 20 terms, we can observe that the series is geometric with the first term \(a = 5\) and common ratio \(r = 3\) (since \(15/5 = 3\)). The \(n\)-th term can be expressed in sigma notation. The series can be written as: \[ x = \sum_{k=0}^{19} 5 \cdot 3^k. \] To find the sum of the first \(n\) terms of a geometric series, we use the formula: \[ S_n = a \frac{(r^n - 1)}{(r - 1)}. \] So for our series: \[ S_{20} = 5 \frac{(3^{20} - 1)}{2} = \frac{5}{2} (3^{20} - 1). \] This yields the final value of \(x\) after evaluating \(3^{20}\).