Merging man & maths https://www.mathcity.org/ Thu, 04 Jun 2026 16:29:22 +0000 FeedCreator 1.8 https://www.mathcity.org/_media/logo.svg https://www.mathcity.org/ https://www.mathcity.org/math-11-kpk/sol/unit05/ex5-4-p1 Question 1 Exercise 5.3 Solutions of Question 1 of Exercise 5.4 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. Question 1(i) $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\ldots$$n$$$T_n=\dfrac{1}{n(n+1)}$$$T_n$$$\dfrac{1}{n(n+1)}=\dfrac{A}{n}+\dfrac{B}{(n+1)}$$$n(n+1)$$$1=A(n+1)+B n=(A+B) n+A$$$n$$$A+B=0 \text{and} A=1$$$A=1$\begin{align}1+B&=0\\ B&=-1\end{align}\beg… anonymous@undisclosed.example.com (Anonymous) Thu, 01 Feb 2024 02:35:20 +0000 https://www.mathcity.org/math-11-kpk/sol/unit05/ex5-4-p2 Question 2 & 3 Exercise 5.4 Solutions of Question 2 & 3 of Exercise 5.4 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 2 $\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2}$\begin{align}\text { Let } S_n&=\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2} \\ S_n&=\sum_{k=1}^n \dfrac{1}{9 k^2+6 k-3 k-2} \\ & =\sum_{k=1}^n \dfrac{1}{3 k(3 k+2)-1(3 k+2)} \\ S_n&=\sum_{k=1}^n \dfrac{1}{(3 k-1… anonymous@undisclosed.example.com (Anonymous) Thu, 01 Feb 2024 02:35:21 +0000 https://www.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p1 Question 1 Exercise 5.1 Solutions of Question 1 of Exercise 5.1 of Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 1(i) $1^2+3^2+5^2+7^2+\ldots$$n$$1+3+5+\ldots$$n^{\text {th }}$$2 n-1$$n^{t h}$$$T_j=(2 j-1)^2$$\begin{align}& \sum_{j=1}^n T_j=\sum_{j=1}^n(2 j-1)^2 \\ & =\sum_{j=1}^n(4 j^2-4 j+1)\\ & =4 \sum_{j=1}^n j^2-4 \sum_{j=1}^n j+\sum_{j=1}^n 1 \\ & =4 \dfr… anonymous@undisclosed.example.com (Anonymous) Thu, 01 Feb 2024 02:35:10 +0000 https://www.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p3 Question 4 & 5 Exercise 5.1 Solutions of Question 4 & 5 of Exercise 5.1 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 4 $2+(2+5)+(2+5+8)+\ldots$$n$\begin{align}& T_j=\dfrac{j}{2}[2(2)+3(j-1)]\\ &=\dfrac{j(3 j+1)}{2} \\ & =\dfrac{1}{2}(3 j^2+j)\end{align}\begin{align}& \sum_{j=1}^n T_i=\dfrac{1}{2}[3 \sum_{j=1}^n j^2+\sum_{j=1}^n j] \\ & =\dfrac{1}{2}[3 \dfra… anonymous@undisclosed.example.com (Anonymous) Thu, 01 Feb 2024 02:35:11 +0000 https://www.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p6 Question 9 Exercise 5.1 Solutions of Question 9 of Exercise 5.1 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 9(i) $n$$n$$n$\begin{align} & T_n=n^2(2 n+3)=2 n^3+3 n^2 \\ & \Rightarrow T_j=2 j^3+3 j^2\end{align}\begin{align} & \sum_{j=1}^n T_j=2 \sum_{j=1}^n j^3+3 \sum_{j=1}^n j^2 \\ & =2(\dfrac{n(n+1)}{2})^2+3 \dfrac{n(n+1)(2 n+1)}{6} \\ & =\dfrac{n(n+1)}{2}… anonymous@undisclosed.example.com (Anonymous) Thu, 01 Feb 2024 02:35:13 +0000 https://www.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p3 Question 4 Review Exercise Solutions of Question 4 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 4 $\dfrac{1}{1.4 .7}+\dfrac{1}{4.7 .10}+\dfrac{1}{7.10 .13}+\ldots$$1,4,7, \ldots$$$a_n=\dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}$$\begin{align} \dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}&=\dfrac{A}{3 n-2}+\dfrac{B}{3 n+1}+\dfrac{C}{3 n+4}\end{align}$(3 n-2)(3 n+… anonymous@undisclosed.example.com (Anonymous) Thu, 01 Feb 2024 02:35:24 +0000 https://www.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p4 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.$5+12 x+19 x^2+26 x^3+\ldots$$n$\begin{align}S_n&=5+12 x+19 x^2+26 x^3+\cdots+(7 n-2) x^{n-1}...(i)\\ x S_n&=5 x+12 x^2+19 x^3+\cdots+(7 n-9) x^{n-1}+(7 n-1) x^n....(ii)\end{align}\begin{align}(1-x) S_n&=5+(12-5) x+(19-12) x^2+\cdots\\ &+[7 n-2-(… anonymous@undisclosed.example.com (Anonymous) Thu, 01 Feb 2024 02:35:24 +0000 https://www.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p4 Question 6 Exercise 5.1 Solutions of Question 6 of Exercise 5.1 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 6 $1.2 \cdot 3+2 \cdot 3.4+3.4 .5+\ldots$$n$$1+2+3+\ldots, \quad 2+3+4+5+\ldots$$3+4+5+6+7+\ldots$$n^{t h}$$j, j+1$$j+2$$n^{t h}$\begin{align} & T_j=j(j+1)(j+2)-j(j^2+3 j+2) \\ & =j^3+3 j^2+2 j\end{align}\begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n … anonymous@undisclosed.example.com (Anonymous) Thu, 01 Feb 2024 02:35:12 +0000 https://www.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p5 Question 7 & 8 Exercise 5.1 Solutions of Question 7 & 8 of Exercise 5.1 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 7 $n$$1.5 .9+2.6 .10+3.7 .11+\ldots$$T_j=j(j+4)(j+8)$\begin{align} & =j(j^2+12 j+32) \\ & =j^3+12 j^2+32 j\end{align}\begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n j^3+12 \sum_{j=1}^n j^2+32 \sum_{j=1}^n j \\ & =(\dfrac{n(n+1)}{2})^2+12 \dfrac… anonymous@undisclosed.example.com (Anonymous) Thu, 01 Feb 2024 02:35:13 +0000 https://www.mathcity.org/math-11-kpk/sol/unit05/ex5-4-p3 Question 4 Exercise 5.4 Solutions of Question 4 of Exercise 5.4 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 4 $\sum_{k=1}^n \dfrac{1}{k^2+7 k+12}$\begin{align}S_n &=\sum_{k=1}^n \dfrac{1}{k^2+7 k+12} \\ & =\sum_{k=1}^n \dfrac{1}{(k+3)(k+4)}\end{align}$n^{\text {th }}$$$u_n=\dfrac{1}{(n+3)(n+4)}$$$$\dfrac{1}{(n+3)(n+4)}=\dfrac{A}{n+3}+\dfrac{B}{n+4}$$$A$$B$… anonymous@undisclosed.example.com (Anonymous) Thu, 01 Feb 2024 02:35:21 +0000