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Question 2 & 3 Exercise 5.1

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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 5.1

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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(