Question Number 29149 by abdo imad last updated on 04/Feb/18
$$\left({u}_{{n}} \right)_{{n}} \:\:{is}\:{arithmetic}\:{progression}/\:{u}_{{n}} =\:{u}_{\mathrm{0}} +{nr}\: \\ $$$${find}\:{S}_{{n}} =\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{u}_{{k}} ^{\mathrm{2}} .\: \\ $$
Answered by mrW2 last updated on 04/Feb/18
![u_k =u_0 +kr u_k ^2 =u_0 ^2 +2u_0 rk+r^2 k^2 Σ_(k=0) ^n u_k ^2 =(n+1)u_0 ^2 +2u_0 rΣ_(k=0) ^n k+r^2 Σ_(k=0) ^n k^2 =(n+1)u_0 ^2 +2u_0 r×(((n+1)n)/2)+r^2 ×((n(n+1)(2n+1))/6) =(n+1)[u_0 ^2 +nu_0 r+((n(2n+1)r^2 )/6)]](https://www.tinkutara.com/question/Q29160.png)
$${u}_{{k}} ={u}_{\mathrm{0}} +{kr} \\ $$$${u}_{{k}} ^{\mathrm{2}} ={u}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{u}_{\mathrm{0}} {rk}+{r}^{\mathrm{2}} {k}^{\mathrm{2}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{u}_{{k}} ^{\mathrm{2}} =\left({n}+\mathrm{1}\right){u}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{u}_{\mathrm{0}} {r}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}+{r}^{\mathrm{2}} \underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \\ $$$$=\left({n}+\mathrm{1}\right){u}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{u}_{\mathrm{0}} {r}×\frac{\left({n}+\mathrm{1}\right){n}}{\mathrm{2}}+{r}^{\mathrm{2}} ×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$=\left({n}+\mathrm{1}\right)\left[{u}_{\mathrm{0}} ^{\mathrm{2}} +{nu}_{\mathrm{0}} {r}+\frac{{n}\left(\mathrm{2}{n}+\mathrm{1}\right){r}^{\mathrm{2}} }{\mathrm{6}}\right] \\ $$