# 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.

Sum the series: $5+12 x+19 x^2+26 x^3+\ldots$ to $n$ terms.

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}

Sum the series: $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\ldots$ to $n$ terms.

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: $$S_n=\dfrac{n}{n+1}$$