This is beta site. https://beta.mathcity.org/ 2026-06-07T21:53:13+00:00 https://beta.mathcity.org/ https://beta.mathcity.org/_media/logo.png text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p1?rev=1737476038&do=diff 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$$\begin{align}& \sum_{j=1}^n T_j=\sum_{j=1}^n(2 j-1)^2 \\ & =\sum_{j=1}^n(4 j^2-4 j+1)\\ & =4 \sum_{j=1}^n j^2-4 \sum_{j=1}^n j+\sum_{j=1}^n 1 \\ & =4 \dfr… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p6?rev=1737476038&do=diff Question 8 Review Exercise Solutions of Question 8 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 8(i) $n$$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$$$S_n=\dfrac{n^2(n+1… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p4?rev=1737476038&do=diff Question 5 & 6 Review Exercise Solutions of Question 5 & 6 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$5+12 x+19 x^2+26 x^3+\ldots$$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-(… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p4?rev=1737476038&do=diff 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$$n^{t h}$\begin{align} & T_j=j(j+1)(j+2)-j(j^2+3 j+2) \\ & =j^3+3 j^2+2 j\end{align}\begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n … text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-1-p5?rev=1737476038&do=diff Question 7 & 8 Exercise 5.1 Solutions of Question 7 & 8 of Exercise 5.1 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 7 $n$$1.5 .9+2.6 .10+3.7 .11+\ldots$$T_j=j(j+4)(j+8)$\begin{align} & =j(j^2+12 j+32) \\ & =j^3+12 j^2+32 j\end{align}\begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n j^3+12 \sum_{j=1}^n j^2+32 \sum_{j=1}^n j \\ & =(\dfrac{n(n+1)}{2})^2+12 \dfrac… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-3-p5?rev=1737476038&do=diff Question 5 Exercise 5.3 Solutions of Question 5 of Exercise 5.3 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 5 $n$$n$$3+9+21+45+93+189+\ldots$\begin{align} & a_2-a_1=9-3=6 \\ & a_3-a_2=21-9=12 \\ & a_4-a_3=45-21=24\\ & \text {... ... ... } \\ & \text {... ... ... } \\ &a_n-a_{n-1}=(\mathrm{n}-1)\quad \text{ term of the sequence}\quad 6,12,24, \ldots\end{ali… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-4-p3?rev=1737476038&do=diff Question 4 Exercise 5.4 Solutions of Question 4 of Exercise 5.4 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 4 $\sum_{k=1}^n \dfrac{1}{k^2+7 k+12}$\begin{align}S_n &=\sum_{k=1}^n \dfrac{1}{k^2+7 k+12} \\ & =\sum_{k=1}^n \dfrac{1}{(k+3)(k+4)}\end{align}$n^{\text {th }}$$$u_n=\dfrac{1}{(n+3)(n+4)}$$$$\dfrac{1}{(n+3)(n+4)}=\dfrac{A}{n+3}+\dfrac{B}{n+4}$$$A$$B$… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-3-p1?rev=1737476038&do=diff Question 1 Exercise 5.3 Solutions of Question 1 of Exercise 5.3 of Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 1 $n$$n$$4+13+28+49+76+\ldots$\begin{align} & a_2-a_1=13-4=9 \\ & a_3-a_2=28-13=15 \\ & a_4-a_3=49-28=21 \\ & \cdots \quad \cdots \quad \cdots \\ & \cdots \quad \cdots \quad \cdots \\ & a_n-a_{n-1}=(n-1)th \quad\text{term of sequence}\quad 9,15,21,..… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-3-p2?rev=1737476038&do=diff Question 2 Exercise 5.3 Solutions of Question 2 of Exercise 5.3 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 2 $n$$n$$4+14+30+52+80+114+\ldots$\begin{align} & a_2-a_1=14-4=10 \\ & a_3-a_2=30-14=16 \\ & a_4-a_3=52-30=22 \\ & \cdots \quad \cdots \quad \cdots \\ & \cdots \quad \cdots \quad \cdots \\ & a_n-a_{n-1}=(\mathrm{n}-1)\text{ term of the sequence} 10,1… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/ex5-3-p3?rev=1737476038&do=diff Question 3 Exercise 5.3 Solutions of Question 3 of Exercise 5.3 of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 3 $n$$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…