Find $\sum\limits_{r=1}^{n}{\left( {{3}^{r}}+16r-8 \right)}$, giving your answer in the form $A\left

Question

Find $\sum\limits_{r=1}^{n}{\left( {{3}^{r}}+16r-8 \right)}$, giving your answer in the form $A\left( {{3}^{n}}-1 \right)+B{{n}^{2}}$, where $A$ and $B$ are constants to be determined. [4] Hence find $\sum\limits_{r=1}^{n}{\left( {{3}^{r+1}}+16r+8 \right)}$ in terms of $n$. [3]

Accepted Answer

(i) $A=\frac{3}{2},\ B=8$ (ii) $\frac{3}{2}(3^{n+1}-1)+8(n+1)^2-11$

2022 CJC J1 MYE Q6

2022 CJC J1 MYE Q6

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