====== Question 11 and 12, Exercise 4.8 ======
Solutions of Question 11 and 12 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 11=====
Evaluate the sum of the series: $\sum_{k=1}^{n} \frac{1}{k(k+2)}$

** Solution. **

Let $T_k$ represent the $k$th term of the series. Then
\begin{align*}
T_k &= \frac{1}{k(k+2)}.
\end{align*}
Resolving it into partial fractions:
\begin{align*}
\frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2} \ldots (1)
\end{align*}
Multiplying both sides by $k(k+2)$, we get
\begin{align*}
	1 = A(k+2) + Bk \ldots (2)
\end{align*}
Put $k=0$ in (2), we have
\begin{align*}
&1=2A + 0 \\
\implies & A = \frac{1}{2}.
\end{align*}
Put $k+2=0 \implies k=-2$ in (2), we have
\begin{align*}
&1=0-2B\\
\implies &B = -\frac{1}{2}.
\end{align*}
Using the values of $A$ and $B$ in equation (1), we get
\begin{align*}
\frac{1}{k(k+2)} &= \frac{1}{2k} - \frac{1}{2(k+2)}.
\end{align*}
Thus,
\begin{align*}
T_k &= \frac{1}{2} \left( \frac{1}{k} - \frac{1}{k+2} \right).
\end{align*}
Taking the sum, we have
\begin{align*}
S_n &= \sum_{k=1}^n T_k = \frac{1}{2} \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+2} \right).
\end{align*}
This is a telescoping series, so most terms cancel out, and we are left with:
\begin{align*}
S_n &= \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right) \\
& = 
\end{align*}
<fc #ff0000>This will be solved later.</fc>




=====Question 12=====
Evaluate the sum of the series: $\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\ldots \ldots$ to infinity.

** Solution. **

Let $T_k$ represents the kth term of the 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 $(3k-2)(3k+1)$
\begin{align*}
1 = (3k+1)A+(3k-2)B   \ldots (2)
\end{align*}
Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have
\begin{align*}
&1 = \left(3\times\frac{2}{3}+1 \right)A+0 \\
\implies &A = \frac{1}{3}. 
\end{align*}
Now put $3k+1=0$ $\implies k=-\dfrac{1}{3}$ in (2), we have
\begin{align*}
&1 = 0+\left(3\left(-\frac{1}{3}\right)-2 \right)B \\
\implies &B = -\frac{1}{3}. 
\end{align*}
Using value of $A$ and $B$ in (1), we get
\begin{align*}
\frac{1}{(3k-2)(3k+1)} = \frac{1}{3(3k-2)}-\frac{1}{3(3k+1)}
\end{align*}
This gives
\begin{align*}
T_k &=\frac{1}{3}\left[\frac{1}{3k-2}-\frac{1}{3k+1}\right]
\end{align*}
Taking sum
\begin{align*}
S_n&=\sum_{k=1}^n T_k =\frac{1}{3}\sum_{k=1}^n\left[\frac{1}{3k-2}-\frac{1}{3k+1}\right] \\
& = \frac{1}{3}\left[\left(\frac{1}{1}-\frac{1}{4}\right) + \left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+\ldots\right.\\
&\left.+\ldots+\left(\frac{1}{3n-5}-\frac{1}{3n-2}\right)+\left(\frac{1}{3n-2}-\frac{1}{3n+1}\right) \right] \\
& = \frac{1}{3}\left[1-\frac{1}{3n+1} \right] \\
& = \frac{1}{3}\left[\frac{3n+1-1}{3n+1} \right] \\
& = \frac{1}{3}\left[\frac{3n}{3n+1} \right] \\
& = \frac{n}{3n+1}
\end{align*}
Now 
\begin{align*}
S_\infty& = \lim_{n \to \infty} S_n \\
&=\lim_{n \to \infty} \frac{n}{3n+1} \\
&=\lim_{n \to \infty} \frac{1}{3+1/n} \\
&=\frac{1}{3+0} = =\frac{1}{3} \\
\end{align*}








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