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- Question 2 & 3 Exercise 5.1
- au_j=\sum_{j=1}^{99} j^2+\sum_{j=1}^{99} j \\ & =\frac{99(99+1)(2(99)+1)}{6}+\frac{99(99+1)}{2} \end{aligned} $$ $$ \begin{aligned} & =\frac{99(100)(199)}{6}+\frac{99(100)}{2} \\ & =(33.50 .199)+(99.50) \\ & \Rightarrow 1.2+2.3+3.4+\ldots+99.10
- Question 1 Exercise 5.3
- =\dfrac{1}{n(n+1)}$$ Resolving $T_n$ into partial fractions $$\dfrac{1}{n(n+1)}=\dfrac{A}{n}+\dfrac{B}{... frac{1}{(2 n-1)(2 n+1)}$. Resolving into partial fractions $$\dfrac{1}{(2 n-1)(2 n-1)}=\dfrac{A}{(2 n-1... dfrac{1}{(3 n-1)(3 n+2)}$$ Resolving into partial fractions: $$\dfrac{1}{(3 n-1)(3 n+2)}=\dfrac{A}{3 n-1... frac{1}{(9 n-5)(9 n+4)}$ Resolving into partial fractions: $$\dfrac{1}{(9 n-5)(9 n+4)}=\dfrac{A}{9 n-5
- Question 2 & 3 Exercise 5.4
- dfrac{1}{(3 k-1)(3 k+2)}$$ Resolving into partial fractions $$\dfrac{1}{(3 k-1)(3 k+2)}=\dfrac{A}{3 k-1}... $$u_n=\dfrac{1}{n(n-1)}$$ Resolving into partial fractions $$\dfrac{1}{k(k-1)}=\dfrac{A}{n}+\dfrac{B}{n
- Question 1 Exercise 5.1
- n+2] \\ & =\dfrac{n(n+1)}{12}[n^2+n+2 n+2]\\ & =\frac{n(n+1)}{12}[n(n+1)+2(n+1)] \\ & =\dfrac{n(n+1)^2(
- Question 4 Exercise 5.4
- _n=\dfrac{1}{(n+3)(n+4)}$$ Resolving into partial fractions $$\dfrac{1}{(n+3)(n+4)}=\dfrac{A}{n+3}+\dfra
- Question 4 Review Exercise
- }{(3 n-2)(3 n+1)(3 n+4)}$$ Resolving into partial fractions \begin{align} \dfrac{1}{(3 n-2)(3 n+1)(3 n+4
- Question 5 & 6 Review Exercise
- =\dfrac{1}{n(n+1)}$$ Resolving $T_n$ into partial fractions $$\dfrac{1}{n(n+1)}=\dfrac{A}{n}+\dfrac{B}{(