Question 9 and 10, Exercise 4.8

Solutions of Question 9 and 10 of Exercise 4.8 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Evaluate the sum of the series: $$\frac{1}{1 \cdot 3}+\frac{1}{2 \cdot 5}+\frac{1}{3 \cdot 7}+\ldots \ldots \text{ up to } \infty$$

Solution.

Do yourself

Evaluate the sum 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} + \frac{B}{k+2} \ldots (1) \end{align*} Multiplying both sides by $(k+1)(k+2)$, we get \begin{align*} 1 = (k+2)A + (k+1)B \ldots (2) \end{align*} Now, put $k+1=0 \implies k=-1$ in equation (2): \begin{align*} 1 &= (-1+2)A + 0 \\ \implies A &= 1. \end{align*} Next, put $k+2=0 \implies k=-2$ in equation (2): \begin{align*} 1 &= 0 + (-2+1)B \\ \implies B &= -1. \end{align*} Using the values of $A$ and $B$ in equation (1), we get \begin{align*} \frac{1}{(k+1)(k+2)} &= \frac{1}{k+1} - \frac{1}{k+2}. \end{align*} Thus, \begin{align*} T_k &= \frac{1}{k+1} - \frac{1}{k+2}. \end{align*} Now \begin{align*} &\sum_{k=3}^{n} \dfrac{1}{(k+1)(k+2)} = \sum_{k=3}^n T_k \\ &= \sum_{k=3}^n \left( \frac{1}{k+1} - \frac{1}{k+2} \right) \\ &= \left( \frac{1}{4} - \frac{1}{5} \right)+ \left( \frac{1}{5} - \frac{1}{6} \right)+\left( \frac{1}{6} - \frac{1}{7} \right) \\ &\quad +...+\left( \frac{1}{n} - \frac{1}{n+1} \right)+ \left( \frac{1}{n+1} - \frac{1}{n+2} \right) \\ &=\frac{1}{4} - \frac{1}{n+2} \\ &=\frac{n+2-4}{4(n+2)} \\ &=\frac{n-2}{4(n+2)} \\ \end{align*} Hence, the required sum $\dfrac{n-2}{4(n+2)}$.

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