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- Question 11 and 12, Exercise 4.8
- ====== Question 11 and 12, Exercise 4.8 ====== Solutions of Question 11 and 12 of Exercise 4.8 of Unit 04: Sequence and Series. ... e sum of the series: $\sum_{k=1}^{n} \frac{1}{k(k+2)}$ ** Solution. ** Let $T_k$ represent the $k$t... e series. Then \begin{align*} T_k &= \frac{1}{k(k+2)}. \end{align*} Resolving it into partial fractio
- Question 13, 14 and 15, Exercise 4.8
- the series. Then \begin{align*} T_k &= \frac{1}{(2k+3)(2k+9)}. \end{align*} Resolving it into partial fractions: \begin{align*} \frac{1}{(2k+3)(2k+9)} = \frac{A}{2k+3} + \frac{B}{2k+9} \ldots (1) \en
- Question 9 and 10, Exercise 4.8
- um of the series: $$\frac{1}{1 \cdot 3}+\frac{1}{2 \cdot 5}+\frac{1}{3 \cdot 7}+\ldots \ldots \text{... of the series: $\sum_{k=3}^{n} \dfrac{1}{(k+1)(k+2)}$ ** Solution. ** Consider \begin{align*} T_k &= \frac{1}{(k+1)(k+2)}. \end{align*} Resolving it into partial fractions: \begin{align*} \frac{1}{(k+1)(k+2)} = \frac{A}{k+1
- Question 7 and 8, Exercise 4.8
- e series. Then \begin{align*} T_k &=\frac{1}{(3k-2)(3k+1)}. \end{align*} Resolving it into partial fraction: \begin{align*} \frac{1}{(3k-2)(3k+1)} = \frac{A}{3k-2}+\frac{B}{3k+1} \ldots (1) \end{align*} Multiplying with $(3k-2)(3k+1)$ \beg