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Question 1 Exercise 5.1

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:10 +0000

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…

Question 2 & 3 Exercise 5.1

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:10 +0000

Question 2 & 3 Exercise 5.1 Solutions of Question 2 & 3 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. Q2 Find the sum $1.2+2.3+3.4+\ldots+99.100$$1+2+3+\ldots+99$$2+3+4+\ldots+100$$n^{\text {th }}$$n(n+1)$$n^{\text {th }}$$\quad T_j=j(j+1)=j^2+j$$j=1$$j=99$$$ \begin{aligned} & \sum_{j=1}^{99} \tau_j=\sum_{j=1}^{99} j^2+\sum_{j=1}^{99} j \\ & =\frac{99…

Question 4 & 5 Exercise 5.1

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:11 +0000

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…

Question 6 Exercise 5.1

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:12 +0000

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 …

Question 7 & 8 Exercise 5.1

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:13 +0000

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…

Question 9 Exercise 5.1

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:13 +0000

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}…

Question 1 Exercise 5.2

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:14 +0000

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. Question 1(i) $n$$1.2+2.2^2+3.2^3+4.2^4+\ldots$\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}\begin{align} (1-2) S_n&=1 \cdot 2+(2-1) 2^2+(3-2) 2^2+(4-…

Question 2 & 3 Exercise 5.2

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:14 +0000

Question 2 & 3 Exercise 5.2 Solutions of Question 2 & 3 of Exercise 5.2 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 $1+3^2 x+5^2 x^2+7^2 x^3+\ldots, x<1$\begin{align} & S_{\infty}=1+3^2 x+5^2 x^2+7^2 x^3+\ldots ..(1)\\ & x S_{\infty}=x+3^2 x^2+5^2 x^3+7^2 x^4+\ldots..(2)\end{align}\begin{align}& (1-x) S_{\infty}=1^2+(3^2-1^2) x+(5^2-3^2) x^2+(7^2-5^2) x^…

Question 4 & 5 Exercise 5.2

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:15 +0000

Question 4 & 5 Exercise 5.2 Solutions of Question 4 & 5 of Exercise 5.2 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 $5+\dfrac{7}{3}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots$\begin{align} & S_{\infty}=5+\dfrac{7}{3}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(i) \\ & \dfrac{1}{3} S_{\infty}=\dfrac{5}{3}+\dfrac{7}{3^2}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(ii) \e…

Question 1 Exercise 5.3

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:16 +0000

Question 1 Exercise 5.3 Solutions of Question 1 of Exercise 5.3 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 $n$$n$$4+13+28+49+76+\ldots$\begin{align} & a_2-a_1=13-4=9 \\ & a_3-a_2=28-13=15 \\ & a_4-a_3=49-28=21 \\ & \cdots \quad \cdots \quad \cdots \\ & \cdots \quad \cdots \quad \cdots \\ & a_n-a_{n-1}=(n-1)th \quad\text{term of sequence}\quad 9,15,21,..…

Question 2 Exercise 5.3

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:17 +0000

Question 2 Exercise 5.3 Solutions of Question 2 of Exercise 5.3 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 $n$$n$$4+14+30+52+80+114+\ldots$\begin{align} & a_2-a_1=14-4=10 \\ & a_3-a_2=30-14=16 \\ & a_4-a_3=52-30=22 \\ & \cdots \quad \cdots \quad \cdots \\ & \cdots \quad \cdots \quad \cdots \\ & a_n-a_{n-1}=(\mathrm{n}-1)\text{ term of the sequence} 10,1…

Question 3 Exercise 5.3

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:17 +0000

Question 3 Exercise 5.3 Solutions of Question 3 of Exercise 5.3 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 3 $n$$n$$4+10+18+28+40+\ldots$\begin{align} & a_2-a_1=10-4=6 \\ & a_3-a_2=18-10=8 \\ & a_4-a_3=28-18=10 \\ & \text {... ... ... } \\ & \text {... ... ... } \\ & a_n-a_{n \quad 1}=(\mathrm{n}-1) \text { term of the sequence } \end{align}$6,10,8, \ldot…

Question 4 Exercise 5.3

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:18 +0000

Question 4 Exercise 5.3 Solutions of Question 4 of Exercise 5.3 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 $n$$n$$3+5+11+29+83+245+\ldots$\begin{align} & a_2-a_1=5-3=2 \\ & a_3-a_2=11-5=6 \\ & a_4-a_3=29-11=18 \\ & \text {... ... ... } \\ & \text {... ... ... } \\ & a_n-a_{n-1}=(\mathrm{n}-1) \text { term ofthe sequence }\end{align}$6,10,18, \ldots$\beg…

Question 5 Exercise 5.3

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:19 +0000

Question 5 Exercise 5.3 Solutions of Question 5 of Exercise 5.3 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 $n$$n$$3+9+21+45+93+189+\ldots$\begin{align} & a_2-a_1=9-3=6 \\ & a_3-a_2=21-9=12 \\ & a_4-a_3=45-21=24\\ & \text {... ... ... } \\ & \text {... ... ... } \\ &a_n-a_{n-1}=(\mathrm{n}-1)\quad \text{ term of the sequence}\quad 6,12,24, \ldots\end{ali…

Question 6 Exercise 5.3

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:20 +0000

Question 6 Exercise 5.3 Solutions of Question 6 of Exercise 5.3 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 $n$$n$$28+32+52+152+652+\ldots$\begin{align} & a_2-a_1=32-28=4 \\ & a_3-a_2=52-32=20 \\ & a_4-a_3=152-52=100 \\ & \ldots \quad \cdots \quad \cdots \\ & \cdots \quad \cdots \quad \cdots \\ & a_n-a_{n-1}=(\mathrm{n}-1) \text { term ofthe sequence } 4…

Question 1 Exercise 5.3

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:20 +0000

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…

Question 2 & 3 Exercise 5.4

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:21 +0000

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…

Question 4 Exercise 5.4

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:21 +0000

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$…

Question 1 Review Exercise 5

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:22 +0000

Question 1 Review Exercise 5 Solutions of Question 1 of Review Exercise 5 of Unit 05: Vectors. 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 $t_n=6 n+5$$t_{n+1}=$$6 n-1$$6 n+11$$6 n+6$$6 n-5$$6 n+11$$1+\dfrac{2}{3}+\dfrac{6}{3^2}+\dfrac{10}{3^3}+\dfrac{14}{3^4}+\ldots$$6$$2$$3$$4$$3$$1+2.2+3.2^2+\cdots+100.2^{\prime \prime}$$99.2^{100}$$100.2^{100}$$99.2^{100}+1$$1000.2^{100}$$99.2^{100}+…

Question 2 & 3 Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:23 +0000

Question 2 & 3 Review Exercise Solutions of Question 2 & 3 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.$n$$1.2+2.3+3.4+\ldots$$n^{\text {th }}$$$a_n=n(n+1)=n^2+n$$\begin{align} \sum_{r=1}^n a_r&=\sum_{r=1}^n r^2+\sum_{r=1}^n r \\ & =\dfrac{n(n+1)(2 n+1)}{6}+\dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)}{2}[\dfrac{2 n+1}{3}+1] \\ & =\dfrac{n(n+1)}{2} \cdot …

Question 4 Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:24 +0000

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+…

Question 5 & 6 Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:24 +0000

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-(…

Question 7 Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:25 +0000

Question 7 Review Exercise Solutions of Question 7 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 7(i) $1.2^2+3.3^2+5.4^2+\ldots$$n$$1,3,5, \ldots,(2 n-1)$$2^2, 3^2, 4^2, \ldots,(n+1)^2$\begin{align} & a_n=(2 n-1)(n+1)^2 \\ & a_n=(2 n-1)(n^2+2 n+1) \\ & a_n=2 n^3+3 n^2-1\end{align}\begin{align} \sum_{r=1}^n a_r&=2 \sum_{r=1}^n r^3+\sum_{r=1…

Question 8 Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:25 +0000

Question 8 Review Exercise Solutions of Question 8 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 8(i) $n$$n^{t h}$$n^3+3^n.$$n^h$$$a_n=n^3+3^n$$\begin{align}\sum_{r=1}^n a_r&=\sum_{r=1}^n r^3+\sum_{r=1}^n 3^r \\ & =[\dfrac{n(n+1)}{2}]^2+\dfrac{3(3^n-1)}{3-1} \\ & =\dfrac{n^2(n+1)^2}{4}+\dfrac{3}{2}(3^n-1) \end{align}$n$$$S_n=\dfrac{n^2(n+1…

Question 9 Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:26 +0000

Question 9 Review Exercise Solutions of Question 9 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 9(i) $n$$3+7+13+21+31+\ldots$\begin{align} & a_2-a_1=7-3=4 \\ & a_3-a_2=13-7=6 \\ & a_4-a_3=21-13=8 \\ & \ldots \quad \ldots \quad \ldots \\ & \ldots \quad \cdots \quad \ldots \\ & a_n-a_{n-1}=(n-1) \text { term of the series } \\ & 4,6,8, \ldo…

Question 10 Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Thu, 01 Feb 2024 02:35:26 +0000

Question 10 Review Exercise Solutions of Question 10 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 10 $n^{\text {th }}$$n$$1+(1+\dfrac{1}{2})+(1+\dfrac{1}{2}+\dfrac{1}{4})+(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8})+\ldots$\begin{align} a_n&=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^{n-1}} \\ a_n&=\dfrac{1[1-(\dfrac{1}{2})…