Question Number 7193 by peter james last updated on 15/Aug/16
Commented by Yozzia last updated on 15/Aug/16
![S(n)=1+2^2 −3^3 +4+5^2 −6^3 +7+8^2 −9^3 +... +[(3n−2)+(3n−1)^2 −(3n)^3 ] S(t=3n)=Σ_(k=1) ^n [(3k−2)+(3k−1)^2 −(3k)^3 ] In grouping the terms of S(n) in threes, t=3n is the total number of terms in the sum−expression of S(n). If we require 3n−1 terms, we can write S(t=3n−1)=Σ_(k=1) ^n [(3k−2)+(3k−1)^2 −(3k)^3 ]+27n^3 where n−1≡ −1 (mod 3) If we require 3n−2 terms, we write S(t=3n−2)=Σ_(k=1) ^n [(3k−2)+(3k−1)^2 −(3k)^3 ]+27n^3 −(3n−1)^2 where n−2≡ −2 (mod 3). So, S(t)= { ((S(t=3n) )),((S(t=3n−1)=S(t=3n)+27n^3 )),((S(t=3n−2)=S(t=3n)+27n^3 −(3n−1)^2 )) :} t=number of terms −−−−−−−−−−−−−−−−−−−−−−−−−− (3k−2)+(3k−1)^2 −27k^3 =3k−2+9k^2 −6k+1−27k^3 =9k^2 −1−3k−27k^3 Using : Σ_(k=1) ^n 1=n , Σ_(k=1) ^n k=((n(n+1))/2) Σ_(k=1) ^n k^2 =((n(n+1)(2n+1))/6) , Σ_(k=1) ^n k^3 =((n^2 (n+1)^2 )/4), S(n)=9×((n(n+1)(2n+1))/6)−n−((3n(n+1))/2)−((27n^2 (n+1)^2 )/4) S(n)=((3n(n+1)(2n+1))/2)−n−((3n(n+1))/2)−((27n^2 (n+1)^2 )/4) S(n)=(n/8)[12(n+1)(2n+1)−8−12(n+1)−54n(n+1)^2 ] S(n)=(n/8)[12(n+1)(2n+1−1)−8−54n(n+1)^2 ] S(n)=(n/8)[24n^2 +24n−8−54n^3 −108n^2 −54n] S(n)=(n/8)[−8−54n^3 −84n^2 −30n] S(n)=−(n/4)(27n^3 +42n^2 +15n+4) n=1⇒S(3)=((−1)/4)(27+42+15+4)=−22 (3 terms) n=2⇒ S(6)=((−1)/2)(27×8+42×4+15×2+4)=−209 (6 terms) t=5⇒S(5)=−209+6^3 =7=1+2^2 −3^3 +4+5^2 −−−−−−−−−−−−−−−−−−−−−−−− So, S(t)= { ((((−n)/4)(27n^3 +42n^2 +15n+4) t≡ 0 (mod 3) )),((((−n)/4)(27n^3 +42n^2 +15n+4)+27n^3 t≡ −1 (mod 3) )),((((−n(27n^3 +42n^2 +15n+4))/4)+27n^3 −(3n−1)^2 t≡ −2 (mod 3) )) :}](Q7195.png)
$${S}\left({n}\right)=\mathrm{1}+\mathrm{2}^{\mathrm{2}} −\mathrm{3}^{\mathrm{3}} +\mathrm{4}+\mathrm{5}^{\mathrm{2}} −\mathrm{6}^{\mathrm{3}} +\mathrm{7}+\mathrm{8}^{\mathrm{2}} −\mathrm{9}^{\mathrm{3}} +... \\ $$$$+\left[\left(\mathrm{3}{n}−\mathrm{2}\right)+\left(\mathrm{3}{n}−\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{3}{n}\right)^{\mathrm{3}} \right] \\ $$$${S}\left({t}=\mathrm{3}{n}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\left(\mathrm{3}{k}−\mathrm{2}\right)+\left(\mathrm{3}{k}−\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{3}{k}\right)^{\mathrm{3}} \right] \\ $$$${In}\:{grouping}\:{the}\:{terms}\:{of}\:{S}\left({n}\right)\:{in}\:{threes}, \\ $$$${t}=\mathrm{3}{n}\:{is}\:{the}\:{total}\:{number}\:{of}\:{terms}\:{in}\:{the} \\ $$$${sum}−{expression}\:{of}\:{S}\left({n}\right).\: \\ $$$${If}\:{we}\:{require}\:\mathrm{3}{n}−\mathrm{1}\:{terms},\:{we}\:{can}\:{write} \\ $$$${S}\left({t}=\mathrm{3}{n}−\mathrm{1}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\left(\mathrm{3}{k}−\mathrm{2}\right)+\left(\mathrm{3}{k}−\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{3}{k}\right)^{\mathrm{3}} \right]+\mathrm{27}{n}^{\mathrm{3}} \\ $$$${where}\:{n}−\mathrm{1}\equiv\:−\mathrm{1}\:\left({mod}\:\mathrm{3}\right) \\ $$$${If}\:{we}\:{require}\:\mathrm{3}{n}−\mathrm{2}\:{terms},\:{we}\:{write} \\ $$$${S}\left({t}=\mathrm{3}{n}−\mathrm{2}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\left(\mathrm{3}{k}−\mathrm{2}\right)+\left(\mathrm{3}{k}−\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{3}{k}\right)^{\mathrm{3}} \right]+\mathrm{27}{n}^{\mathrm{3}} −\left(\mathrm{3}{n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${where}\:{n}−\mathrm{2}\equiv\:−\mathrm{2}\:\left({mod}\:\mathrm{3}\right). \\ $$$${So},\:{S}\left({t}\right)=\begin{cases}{{S}\left({t}=\mathrm{3}{n}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\\{{S}\left({t}=\mathrm{3}{n}−\mathrm{1}\right)={S}\left({t}=\mathrm{3}{n}\right)+\mathrm{27}{n}^{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\\{{S}\left({t}=\mathrm{3}{n}−\mathrm{2}\right)={S}\left({t}=\mathrm{3}{n}\right)+\mathrm{27}{n}^{\mathrm{3}} −\left(\mathrm{3}{n}−\mathrm{1}\right)^{\mathrm{2}} \:\:}\end{cases} \\ $$$${t}={number}\:{of}\:{terms} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\left(\mathrm{3}{k}−\mathrm{2}\right)+\left(\mathrm{3}{k}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{27}{k}^{\mathrm{3}} \\ $$$$=\mathrm{3}{k}−\mathrm{2}+\mathrm{9}{k}^{\mathrm{2}} −\mathrm{6}{k}+\mathrm{1}−\mathrm{27}{k}^{\mathrm{3}} \\ $$$$=\mathrm{9}{k}^{\mathrm{2}} −\mathrm{1}−\mathrm{3}{k}−\mathrm{27}{k}^{\mathrm{3}} \\ $$$${Using}\:: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1}={n}\:,\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\:,\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} =\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}, \\ $$$${S}\left({n}\right)=\mathrm{9}×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−{n}−\frac{\mathrm{3}{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{27}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$${S}\left({n}\right)=\frac{\mathrm{3}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}}−{n}−\frac{\mathrm{3}{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{27}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$${S}\left({n}\right)=\frac{{n}}{\mathrm{8}}\left[\mathrm{12}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{8}−\mathrm{12}\left({n}+\mathrm{1}\right)−\mathrm{54}{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} \right] \\ $$$${S}\left({n}\right)=\frac{{n}}{\mathrm{8}}\left[\mathrm{12}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}−\mathrm{1}\right)−\mathrm{8}−\mathrm{54}{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} \right] \\ $$$${S}\left({n}\right)=\frac{{n}}{\mathrm{8}}\left[\mathrm{24}{n}^{\mathrm{2}} +\mathrm{24}{n}−\mathrm{8}−\mathrm{54}{n}^{\mathrm{3}} −\mathrm{108}{n}^{\mathrm{2}} −\mathrm{54}{n}\right] \\ $$$${S}\left({n}\right)=\frac{{n}}{\mathrm{8}}\left[−\mathrm{8}−\mathrm{54}{n}^{\mathrm{3}} −\mathrm{84}{n}^{\mathrm{2}} −\mathrm{30}{n}\right] \\ $$$${S}\left({n}\right)=−\frac{{n}}{\mathrm{4}}\left(\mathrm{27}{n}^{\mathrm{3}} +\mathrm{42}{n}^{\mathrm{2}} +\mathrm{15}{n}+\mathrm{4}\right) \\ $$$${n}=\mathrm{1}\Rightarrow{S}\left(\mathrm{3}\right)=\frac{−\mathrm{1}}{\mathrm{4}}\left(\mathrm{27}+\mathrm{42}+\mathrm{15}+\mathrm{4}\right)=−\mathrm{22}\:\left(\mathrm{3}\:{terms}\right) \\ $$$${n}=\mathrm{2}\Rightarrow\:{S}\left(\mathrm{6}\right)=\frac{−\mathrm{1}}{\mathrm{2}}\left(\mathrm{27}×\mathrm{8}+\mathrm{42}×\mathrm{4}+\mathrm{15}×\mathrm{2}+\mathrm{4}\right)=−\mathrm{209}\:\:\left(\mathrm{6}\:{terms}\right) \\ $$$${t}=\mathrm{5}\Rightarrow{S}\left(\mathrm{5}\right)=−\mathrm{209}+\mathrm{6}^{\mathrm{3}} =\mathrm{7}=\mathrm{1}+\mathrm{2}^{\mathrm{2}} −\mathrm{3}^{\mathrm{3}} +\mathrm{4}+\mathrm{5}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${So},\:{S}\left({t}\right)=\begin{cases}{\frac{−{n}}{\mathrm{4}}\left(\mathrm{27}{n}^{\mathrm{3}} +\mathrm{42}{n}^{\mathrm{2}} +\mathrm{15}{n}+\mathrm{4}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}\equiv\:\mathrm{0}\:\left({mod}\:\mathrm{3}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\\{\frac{−{n}}{\mathrm{4}}\left(\mathrm{27}{n}^{\mathrm{3}} +\mathrm{42}{n}^{\mathrm{2}} +\mathrm{15}{n}+\mathrm{4}\right)+\mathrm{27}{n}^{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}\equiv\:−\mathrm{1}\:\left({mod}\:\mathrm{3}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\\{\frac{−{n}\left(\mathrm{27}{n}^{\mathrm{3}} +\mathrm{42}{n}^{\mathrm{2}} +\mathrm{15}{n}+\mathrm{4}\right)}{\mathrm{4}}+\mathrm{27}{n}^{\mathrm{3}} −\left(\mathrm{3}{n}−\mathrm{1}\right)^{\mathrm{2}} \:\:\:\:{t}\equiv\:−\mathrm{2}\:\left({mod}\:\mathrm{3}\right)\:\:}\end{cases} \\ $$$$ \\ $$
Commented by peter james last updated on 16/Aug/16
$$\boldsymbol{{T}}{hank}\:{you}\:{so}\:{much}\:{for}\:{the}\:{help}. \\ $$