2022 SAJC Promo Q2
2022 SAJC Promo Q2
12 marks
Promo
- Using the method of difference, show that \(\sum\limits_{r=\,\,1}^{n}{\left[ {{\left( r+1 \right)}^{3}}-{{r}^{3}} \right]}={{\left( n+1 \right)}^{3}}-1\).[2]
- Expand and simplify \({{\left( r+1 \right)}^{3}}-{{r}^{3}}\).[1]
- Hence, by using (i) and (ii), show that\(\sum\limits_{r=\,\,1}^{n}{{{r}^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}\).[3]
- It is given that \(\sum\limits_{r=\,\,1}^{n}{\frac{2r+1}{r\left( r+1 \right)\left( r+2 \right)}=\frac{n\left( 5n+7 \right)}{4\left( n+1 \right)\left( n+2 \right)}}\).
- Show that the series \(\sum\limits_{r=\,\,1}^{\infty }{\frac{2r+1}{r\left( r+1 \right)\left( r+2 \right)}}\) converges and write down its sum to infinity.[3]
- Find \(\sum\limits_{r=\,0}^{n-2}{\frac{2r+5}{\left( r+2 \right)\left( r+3 \right)\left( r+4 \right)}}\).[3]
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Answer:(a)(ii) \(3{{r}^{2}}+3r+1\) (b)(ii) \(\frac{n\left( 5n+7 \right)}{4\left( n+1 \right)\left( n+2 \right)}-\frac{1}{2}\)
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