Question Number 114433 by bemath last updated on 19/Sep/20
$${find}\:{sum}\:{of}\:{the}\:{series}\: \\ $$$$\mathrm{1}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} −\mathrm{11}^{\mathrm{2}} +…+\left(\mathrm{4}{n}−\mathrm{3}\right)^{\mathrm{2}} −\left(\mathrm{4}{n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$
Answered by 1549442205PVT last updated on 19/Sep/20
![S=1+(5^2 −3^2 )+(9^2 −7^2 )+(13^2 −11^2 )+...[(4n−3)^2 −(4n−5)^2 ]−(4n−1)^2 =1+(5+3)(5−3)+(9+7)(9−7)+(13+11)(13−11)+...+(8n−8).2−(4n−1)^2 =1+2[8+16+24+...+(8n−8)]−(4n−1)^2 =1+2[(((8+8n−8)(n−1))/2)]−(4n−1)^2 =1+8n(n−1)−16n^2 +8n−1 S=−8n^2](https://www.tinkutara.com/question/Q114439.png)
$$\mathrm{S}=\mathrm{1}+\left(\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} \right)+\left(\mathrm{9}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} \right)+\left(\mathrm{13}^{\mathrm{2}} −\mathrm{11}^{\mathrm{2}} \right)+…\left[\left(\mathrm{4n}−\mathrm{3}\right)^{\mathrm{2}} −\left(\mathrm{4n}−\mathrm{5}\right)^{\mathrm{2}} \right]−\left(\mathrm{4n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}+\left(\mathrm{5}+\mathrm{3}\right)\left(\mathrm{5}−\mathrm{3}\right)+\left(\mathrm{9}+\mathrm{7}\right)\left(\mathrm{9}−\mathrm{7}\right)+\left(\mathrm{13}+\mathrm{11}\right)\left(\mathrm{13}−\mathrm{11}\right)+…+\left(\mathrm{8n}−\mathrm{8}\right).\mathrm{2}−\left(\mathrm{4n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}+\mathrm{2}\left[\mathrm{8}+\mathrm{16}+\mathrm{24}+…+\left(\mathrm{8n}−\mathrm{8}\right)\right]−\left(\mathrm{4n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}+\mathrm{2}\left[\frac{\left(\mathrm{8}+\mathrm{8n}−\mathrm{8}\right)\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}}\right]−\left(\mathrm{4n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}+\mathrm{8n}\left(\mathrm{n}−\mathrm{1}\right)−\mathrm{16n}^{\mathrm{2}} +\mathrm{8n}−\mathrm{1} \\ $$$$\mathrm{S}=−\mathrm{8n}^{\mathrm{2}} \\ $$
Commented by bemath last updated on 19/Sep/20
$${yes}..{gave}\:{kudos}\:\checkmark\: \\ $$