====== Question 5 & 6 Review Exercise ======
Solutions of Question 5 & 6 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 5=====
Sum the series: $5+12 x+19 x^2+26 x^3+\ldots$ to $n$ terms.
====Solution====
Let
\begin{align}S_n&=5+12 x+19 x^2+26 x^3+\cdots+(7 n-2) x^{n-1}...(i)\\ 
x S_n&=5 x+12 x^2+19 x^3+\cdots+(7 n-9) x^{n-1}+(7 n-1) x^n....(ii)\end{align}
Subtracting the (ii) from (i) we get
\begin{align}(1-x) S_n&=5+(12-5) x+(19-12) x^2+\cdots\\
&+[7 n-2-(7 n-9)] x^{n-1}-(7 n-1) x^n \\
& =5+7 x+7 x^2+\cdots+7 x^{n-1}-(7 n-1) x^n \\
\Rightarrow(1-x) S_n&=5+7[x+x^2+\cdots+ x^{n-1}]-(7 n-1) x^n \\
\Rightarrow(1-x) S_n&=5+7 \cdot \dfrac{x(1-x^{n-1})}{1-x}-(7 n-1) x^n \\
\Rightarrow S_n&=\dfrac{5}{1-x}+\dfrac{7(x-x^n)}{(1-x)^2}-\dfrac{(7 n-1) x^n}{1-x}\\
\end{align}

=====Question 6=====
Sum the series: $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\ldots$ to $n$ terms.
====Solution====
Solution: The general term of the series is:
$$T_n=\dfrac{1}{n(n+1)}$$
Resolving $T_n$ into partial fractions
$$\dfrac{1}{n(n+1)}=\dfrac{A}{n}+\dfrac{B}{(n+1)}$$
Multiplying both sides by $n(n+1)$, we get
$$1=A(n+1)+B n=(A+B) n+A$$
Comparing the coefficients of $n$ and constants on the both sides of the above equation, we get
$$A+B=0\quad \text{and}\quad A=1$$
Putting $A=1$ in the $1+B=0$,
we get $$B=-1$$
\begin{align}\dfrac{1}{n(n+1)}&=\dfrac{1}{n}-\dfrac{1}{n+1}\\
T_n&=\dfrac{1}{n}-\dfrac{1}{n+1}\end{align}
Taking summation of the both sides of the above equation
\begin{align}
\sum_{k=1}^n T_k&=\sum_{k=1}^n(\dfrac{1}{k}-\dfrac{1}{k+1}) \\
& =(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+(\dfrac{1}{3}-\dfrac{1}{4})+ \\
& \cdots+(\dfrac{1}{n}-\dfrac{1}{n+1}) \\
& =1-\dfrac{1}{n+1}=\dfrac{n}{n+1} \end{align}
Hence the sum of the series is:
$$S_n=\dfrac{n}{n+1} $$


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