This is beta site. https://beta.mathcity.org/ 2026-06-07T03:25:32+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-p2?rev=1737476038&do=diff 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$$$ \begin{aligned} & \sum_{j=1}^{99} \tau_j=\sum_{j=1}^{99} j^2+\sum_{j=1}^{99} j \\ & =\frac{99… text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p2?rev=1737476038&do=diff Question 2 & 3 Review Exercise Solutions of Question 2 & 3 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$n$$1.2+2.3+3.4+\ldots$$n^{\text {th }}$$$a_n=n(n+1)=n^2+n$$\begin{align} \sum_{r=1}^n a_r&=\sum_{r=1}^n r^2+\sum_{r=1}^n r \\ & =\dfrac{n(n+1)(2 n+1)}{6}+\dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)}{2}[\dfrac{2 n+1}{3}+1] \\ & =\dfrac{n(n+1)}{2} \cdot … text/html 2025-01-21T16:13:58+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-kpk/sol/unit05/re-ex5-p3?rev=1737476038&do=diff Question 4 Review Exercise Solutions of Question 4 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan. Question 4 $\dfrac{1}{1.4 .7}+\dfrac{1}{4.7 .10}+\dfrac{1}{7.10 .13}+\ldots$$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+…