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Question 2 & 3 Exercise 5.1

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t {th }}$ terms are $n(n+1)$ and the given series have 99 terms. Therefore, the $n^{\text {th }}$ term o

Question 2 & 3 Review Exercise

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dots, 3,4,5 \ldots$ and $5,6,7, \ldots$ Each one have $n^{t h}$ term $n, \quad n+2$ and $n+4$ respectiv

Question 4 Review Exercise

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, B=-\dfrac{1}{9}$ and $C=\dfrac{1}{18}$. Thus we have \begin{align} & a_n=\dfrac{1}{18(3 n-2)}-\dfrac{1