Question Number 109901 by ajfour last updated on 26/Aug/20
$${If}\:{first}\:{n}\:{terms}\:{of}\:{an}\:{A}.{P}.\:{is}\:{cn}^{\mathrm{2}} , \\ $$$${find}\:{sum}\:{of}\:{squares}\:{of}\:{its}\:{first} \\ $$$${n}\:{terms}. \\ $$
Answered by mr W last updated on 26/Aug/20
![a_1 +(a_1 +d)+(a_1 +2d)+....+(a_1 +(n−1)d) =na_1 +((n(n−1)d)/2)=n[a_1 −(d/2)+((nd)/2)]=cn^2 ⇒a_1 −(d/2)=0 ⇒d=2a_1 ⇒(d/2)=c ⇒d=2c ⇒a_1 =c, d=2c c^2 +(3c)^2 +(5c)^2 +...((2n−1)c)^2 =[1^2 +3^2 +5^2 +...(2n−1)^2 ]c^2 =((n(4n^2 −1)c^2 )/3)](https://www.tinkutara.com/question/Q109904.png)
$${a}_{\mathrm{1}} +\left({a}_{\mathrm{1}} +{d}\right)+\left({a}_{\mathrm{1}} +\mathrm{2}{d}\right)+….+\left({a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}\right) \\ $$$$={na}_{\mathrm{1}} +\frac{{n}\left({n}−\mathrm{1}\right){d}}{\mathrm{2}}={n}\left[{a}_{\mathrm{1}} −\frac{{d}}{\mathrm{2}}+\frac{{nd}}{\mathrm{2}}\right]={cn}^{\mathrm{2}} \\ $$$$\Rightarrow{a}_{\mathrm{1}} −\frac{{d}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow{d}=\mathrm{2}{a}_{\mathrm{1}} \\ $$$$\Rightarrow\frac{{d}}{\mathrm{2}}={c}\:\Rightarrow{d}=\mathrm{2}{c} \\ $$$$\Rightarrow{a}_{\mathrm{1}} ={c},\:{d}=\mathrm{2}{c} \\ $$$$ \\ $$$${c}^{\mathrm{2}} +\left(\mathrm{3}{c}\right)^{\mathrm{2}} +\left(\mathrm{5}{c}\right)^{\mathrm{2}} +…\left(\left(\mathrm{2}{n}−\mathrm{1}\right){c}\right)^{\mathrm{2}} \\ $$$$=\left[\mathrm{1}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +…\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} \right]{c}^{\mathrm{2}} \\ $$$$=\frac{{n}\left(\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}\right){c}^{\mathrm{2}} }{\mathrm{3}} \\ $$
Commented by ajfour last updated on 26/Aug/20
$${Marvellous}\:{Sir},\:{too}\:{good}! \\ $$$${but}\:{how}\:{could}\:{you}\:{know}\:{the} \\ $$$${answer}\:{after}\:{the}\:\mathrm{2}^{{nd}} \:{last}\:{step}, \\ $$$${Sir}? \\ $$
Commented by mr W last updated on 26/Aug/20
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{4}{k}^{\mathrm{2}} −\mathrm{4}{k}+\mathrm{1}\right) \\ $$$$=\mathrm{4}×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−\mathrm{4}×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+{n} \\ $$$$=\frac{\mathrm{4}\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)}{\mathrm{3}}+{n} \\ $$$$=\left(\frac{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{4}+\mathrm{3}}{\mathrm{3}}\right){n} \\ $$$$=\frac{{n}\left(\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{3}} \\ $$