Question 1 Exercise 5.2

Solutions of Question 1 of Exercise 5.2 of Unit 05: Mascellaneous 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 up to $n$ terms the series $1.2+2.2^2+3.2^3+4.2^4+\ldots$.

Let $$\begin{align} & S_n=1.2+2.2^2+3 \cdot 2^3+4 \cdot 2^4+\ldots +n \cdot 2^n....(i) \\ & 2 S_n=1.2^2+2.2^3+3.2^4+4.2^5+\ldots +n \cdot 2^n....(ii)\end{align}$$ Suburacting the (ii) from (i), we get $$\begin{align} (1-2) S_n&=1 \cdot 2+(2-1) 2^2+(3-2) 2^2+(4-3) 2^3+\ldots \\ & +(n-(n-1)) 2^n-n \cdot 2^{n+1} \\ & =(2+2^2+2^3++\ldots+2^n)-n \cdot 2^{n+1} \\ & =\dfrac{2(2^n-1)}{2-1}-n \cdot 2^{n+1} \\ & =2^{n+1}-2-n \cdot 2^{n+1} \\ & =2+n \cdot 2^{n+1}-2^{n+1} \\ S_n& =2+(n-1) 2^{n+1}\end{align}$$

Sum up to $n$ terms the series $1+4 x+7 x^2+10 x^3+\ldots.$

Let $$\begin{align} & S_n=1+4 x+7 x^2+10 x^3+\ldots +(3 n-2) x^{n-1}....(i) \\ & x S_n=x+4 x^2+7 x^3+10 x^4+\ldots +(3 n-2) x^{4 t}....(ii)\end{align}$$ Subtracting the (ii) from (i), we get $$\begin{align} (1-x) S_n&=1+(4-1) x+(7-4) x^2+ (10-7) x^3+\ldots \\ & +(3 n-2,3 n-5)) x n-1-(3 n-2) x^n \\ & =1+3 x+3 x^2+3 x^3+\ldots +3 x^{n-1}-(3 n-2) x^n \\ & =1+\dfrac{3 x(1-x^{n-1})}{1-x}-(3 n-2) x^n \\ S_n&=\dfrac{1-(3 n-2) x^n}{1-x}+\dfrac{3 x(1-x^{n-1})}{(1-x)^2}\end{align}$$

Sum up to $n$ terms the series $1+2 x+3 x^2+4 x^3+\ldots$.

Let $$\begin{align} & S_n=1+2 x+3 x^2+4 x^3+\ldots +(n-1)^n 2+n x^n....(i) \\ & x S_n=x+2 x^2+3 x^3+4 x^4+\ldots +(n-1)^{n-1}+n x^n....(ii) \end{align}$$ Subtracting the (ii) from (i), we get $$\begin{align}& (1-x) S_n=1+(2-1) x+(3-2) x^2 \\ & +(4-3) x^3+\ldots +(n-(n-1)) x^{n-1}-n x^n \\ & =1+x+x^2+x^3+\ldots+x^{n-1}-n x^n \\ & =\dfrac{1(1-x^n)}{1-x}-n x^n \\ S_n& =\dfrac{1-x^n}{(1-x)^2}-\dfrac{n x^n}{1-x} \end{align}$$

Sum up to $n$ terms the series $1+\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{7}{8}+\ldots$

Let $$\begin{align} & S_n=1+\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{7}{8}+\ldots+\dfrac{2 n-1}{2^{n-1}} \\ & S_n=1+\dfrac{3}{2}+\dfrac{5}{2^2}+\dfrac{7}{2^3}+\ldots+\dfrac{2 n-1}{2^{n-1}}....(i)\end{align}$$ Multiply (i) both sides by $\dfrac{1}{2}$, we get $$\dfrac{1}{2} S_n=\dfrac{1}{2}+\dfrac{3}{2^2}+\dfrac{5}{2^3}+\dfrac{7}{2^4}+\ldots+\dfrac{2 n-1}{2^n}....(ii)$$ Subtracting the (ii) from (i), we get $$\begin{align} &(1-\dfrac{1}{2}) S_n=1+(\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+\ldots \\ & +\dfrac{2}{2^{n-1}})-\dfrac{2 n-1}{2^n}\\ & =1+(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\ldots+\dfrac{1}{2^{n-1}})-\dfrac{2 n-1}{2^n} \\ &\dfrac{1}{2}S_n =1+\dfrac{1(1-(\dfrac{1}{2})^{n -1})}{1-\dfrac{1}{2}}-\dfrac{2 n-1}{2^n} \\ &S_n =2+4[1-(\dfrac{1}{2})^{n-1}]-2\dfrac{2 n-1}{2^n} \\ &S_n =2+4[1-(\dfrac{1}{2})^{n-1}]-\dfrac{2 n-1}{2^{n-1}}\end{align}$$

Sum up to $n$ terms the series $1-7 x+13 x^2-19 x^3+\ldots$

Let $$\begin{align}& S_n=1+7(-x)+13(-x)^2+19(-x)^3+\ldots-(6 n+5)(-x)^{n-1}....(i) \\ & -x S_n=-x+7(-x)^2+13(-x)^3+19(-x)^4+\ldots+(6 n+5)(-x)^n....(ii) \end{align}$$ Subtracting the (ii) from (i), we get $$\begin{align} & (1+x) S_n=1+(7-1)(-x)+(13- 7)(-x)^2+\ldots+[6 n+5-(6 n-1)]\times(-x)^{n-1}-(6 n+5)(-x)^n \\ & =1+6(-x)+6(-x)^2+\ldots+6(-x)^{n-1}-(6 n+5)(-x)^n \\ & =1+6\dfrac{(-x)(1-(-x)^{n-1})}{1+x}-(6 n+5)(-x)^n \\ & \Rightarrow S_n =\dfrac{1-(6 n+5)(-x)^n}{1+x}-6 x\dfrac{(1-(-x))^{n-1}}{(1+x)^2}\end{align}$$

Question 2 & 3 >