1 Consider the sequence 21, \( , 36, b, 55 \) which has a second constant second difference of 1 11 Determine the values of \( a \) and \( b \) 12 Determine the \( \pi^{\text {to }} \) term of the sequence. (16) I. 3 Hence prove that the sum of any two consecutive terms in the sequence is a square number. ( \( S \) ) 2 Consider the sequence \( \frac{1}{2}+\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)+\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+\ldots+\left(\frac{1}{200}+\frac{2}{200}+\frac{3}{200}+\ldots+\frac{199}{200}\right) \) with \( \Gamma_{2}=\frac{1}{2} \) and \( T_{2}=\left(\frac{1}{3}+\frac{2}{3}\right)=1 \) and \( T_{n}=\left(\frac{1}{200}+\frac{2}{200}+\frac{3}{200}+\ldots \frac{199}{200}\right) \) 21 Calculate the values of the third and fourth terms. 2.2 Determine the number of terms in the sequence. 2.3 What is the value of the last term in the sequence? 24 Calculate the sum of all the terms in the sequence. Prove that: \( \sum_{n=1}^{k} 2^{n-1}=2^{k}-1 \) Prove, without using a formula, that the sum of the series \( 3^{2}+3^{4}+3^{6}+\ldots \) for twenty terms

To solve the sequence problem, we start with the given terms: 21, \( a \), 36, \( b \), 55. Since the second differences are constant and equal to 1, this suggests a quadratic sequence. By calculating the first differences, you’ll find them to be the sequence of numbers increasing by 2 each time. From this, you can deduce the values of \( a \) (which is 28) and \( b \) (which is 45). Now, for the \( n^{\text{th}} \) term, you can derive the formula based on the quadratic nature, yielding \( a_n = \frac{1}{2}n^2 + \frac{41}{2}n - 7 \). To show the sum of any two consecutive terms is a perfect square, you can express the sum as \( a_n + a_{n+1} \) and demonstrate this results in a perfect square through proper algebraic manipulation. The sequence you’re examining starts with \( \frac{1}{2} \) and continues to \( \frac{199}{200} \). The terms of the form \( \frac{1}{n} + \frac{2}{n} + ... + \frac{(n-1)}{n} \) sum up in a controlled manner. For the last term in this sequence, which is \( T_{200} \), you’ll find the sum simplifies nicely to \( \frac{199 \times 200}{2 \times 200} = \frac{199}{2} \). For the series sum \( \sum_{n=1}^{k} 2^{n-1} = 2^{k} - 1 \), one can prove it inductively: If it holds for \( n \), show it holds for \( n + 1 \) by manipulating the left side to fit the suggested form. Similarly, the series \( 3^2 + 3^4 + ... + 3^{20} \) can be computed as a geometric progression where the first term is \( 3^2 \), the ratio is \( 3^2 \), and it can be summed to reach a neat conclusion about its total values over twenty terms.