====== 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. 

=====Question 13=====
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*}
<fc #ff0000>The solution seems very lengthy, it will be solved later.</fc>




=====Question 14=====
Evaluate the sum of the series: $\sum_{k=1}^{n} \frac{1}{9 k^{2}+2 k-2}$

** Solution. **




=====Question 15=====
Evaluate the sum of the series: $\sum_{k=2}^{n} \frac{1}{k^{2}-k}$

** Solution. **









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<text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-8-p6|< Question 11 & 12]]</btn></text>