Question Number 189248 by 073 last updated on 13/Mar/23
Answered by mr W last updated on 13/Mar/23
![Σ_(n=1) ^(1012) [(2n−1)^3 −(2n)^3 ] =Σ_(n=1) ^(1012) [−12n^2 +6n−1] =−12×((1012×1013×(2×1012+1))/6)+6×((1012×1013)/2)−1012 =−4051×1012^2 ⇒answer (B)](https://www.tinkutara.com/question/Q189251.png)
$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{1012}} {\sum}}\left[\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} −\left(\mathrm{2}{n}\right)^{\mathrm{3}} \right] \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{1012}} {\sum}}\left[−\mathrm{12}{n}^{\mathrm{2}} +\mathrm{6}{n}−\mathrm{1}\right] \\ $$$$=−\mathrm{12}×\frac{\mathrm{1012}×\mathrm{1013}×\left(\mathrm{2}×\mathrm{1012}+\mathrm{1}\right)}{\mathrm{6}}+\mathrm{6}×\frac{\mathrm{1012}×\mathrm{1013}}{\mathrm{2}}−\mathrm{1012} \\ $$$$=−\mathrm{4051}×\mathrm{1012}^{\mathrm{2}} \\ $$$$\Rightarrow{answer}\:\left({B}\right) \\ $$
Commented by 073 last updated on 13/Mar/23
$$\mathrm{nice}\:\mathrm{solution} \\ $$
Answered by mehdee42 last updated on 14/Mar/23
$${A}=−\left[\left(\mathrm{1}^{\mathrm{2}} +\mathrm{1}×\mathrm{2}+\mathrm{2}^{\mathrm{2}} \right)+\left(\mathrm{3}^{\mathrm{2}} +\mathrm{3}×\mathrm{4}+\mathrm{4}^{\mathrm{2}} \right)+…+\left(\mathrm{2023}^{\mathrm{2}} +\mathrm{2023}×\mathrm{2024}+\mathrm{2024}^{\mathrm{2}} \right)\right. \\ $$$$=−\left(\underset{\mathrm{1}} {\overset{\mathrm{2024}} {\sum}}{i}^{\mathrm{2}} +\underset{\mathrm{1}} {\overset{\mathrm{1012}} {\sum}}\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}\right)\right) \\ $$$$=−\mathrm{1012}^{\mathrm{2}} ×\mathrm{4051} \\ $$