Question 13, 14 and 15, Exercise 4.8

Solutions of Question 13, 14 and 15 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: $\frac{1}{5 \cdot 11}+\frac{1}{7 \cdot 13}+\frac{1}{9 \cdot 15}+\ldots \ldots$ to $n$ term.

Solution. Let $T_k$ represent the $k$th term of the series. Then $$\begin{align*} T_k &= \frac{1}{(2k+3)(2k+9)}. \end{align*}$$ Resolving it into partial fractions: $$\begin{align*} \frac{1}{(2k+3)(2k+9)} = \frac{A}{2k+3} + \frac{B}{2k+9} \ldots (1) \end{align*}$$ Multiplying both sides by $(2k+3)(2k+9)$, we get $$\begin{align*} 1 = (2k+9)A + (2k+3)B \ldots (2) \end{align*}$$ Now, put $2k+3 = 0 \implies k = -\frac{3}{2}$ in equation (2): $$\begin{align*} 1 &= (2 \times \left(-\frac{3}{2}\right)+9)A + 0 \\ 1 &= 6A \\ \implies A &= \frac{1}{6}. \end{align*}$$ Next, put $2k+9 = 0 \implies k = -\frac{9}{2}$ in equation (2): $$\begin{align*} 1 &= 0 + (2 \times \left(-\frac{9}{2}\right)+3)B \\ 1 &= (-9+3)B \\ 1 &= -6B \\ \implies B &= -\frac{1}{6}. \end{align*}$$ Using the values of $A$ and $B$ in equation (1), we get $$\begin{align*} \frac{1}{(2k+3)(2k+9)} &= \frac{1}{6(2k+3)} - \frac{1}{6(2k+9)}. \end{align*}$$ Thus, $$\begin{align*} T_k &= \frac{1}{6} \left( \frac{1}{2k+3} - \frac{1}{2k+9} \right). \end{align*}$$ Taking the sum, we have $$\begin{align*} S_n &= \sum_{k=1}^n T_k = \frac{1}{6} \sum_{k=1}^n \left( \frac{1}{2k+3} - \frac{1}{2k+9} \right). \end{align*}$$ The solution seems very lengthy, it will be solved later.

Evaluate the sum of the series: $\sum_{k=1}^{n} \frac{1}{9 k^{2}+2 k-2}$

Solution.

Evaluate the sum of the series: $\sum_{k=2}^{n} \frac{1}{k^{2}-k}$

Solution.

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