Menu Close

Question-186657

Tinku Tara 0 Comments
Pin




Question Number 186657 by Mingma last updated on 08/Feb/23
Answered by Rasheed.Sindhi last updated on 08/Feb/23
• Σ_(k=1) ^n T_k : determinant (((T_(n+1) −T_n ),a_n ),((T_2 −T_1 ),a_1 ),((T_3 −T_2 ),a_2 ),((T_4 −T_3 ),a_3 ),((....),(...)),((T_n −T_(n−1) ),a_(n−1) ),((T_(n+1) −T_n ),a_n ),((T_(n+1) −T_1 ),((n/2){2a_1 +(n−1)d}))) T_(n+1) −T_1 =(n/2){2(7)+8(n−1)} T_(n+1) −T_1 =n{7+4(n−1)} T_(n+1) −3=n(4n+3) T_(n+1) =4n^2 +3n+3 T_n =4(n−1)^2 +3(n−1)+3 =4(n^2 −2n+1)+3n−3+3 =4n^2 −8n+4+3n =4n^2 −5n+4 determinant (((T_n =4n^2 −5n+4))) Σ_(k=1) ^n T_k =4Σ_(k=1) ^(n) k^2 −5Σ_(k=1) ^(n) k+4Σ_(k=1) ^(n) 1 =4(((n(n+1)(2n+1))/6))−5(((n(n+1))/2))+4n =((2n(n+1)(2n+1))/3)−((5n(n+1))/2)+4n =((4n(n+1)(2n+1)−15n(n+1)+24n)/6) =n(((8n^2 +12n+4−15n−15+24)/6)) =n(((8n^2 −3n+13)/6)) =((n(8n^2 −3n+13))/6) determinant (((Σ_(k=1) ^n T_k =((n(8n^2 −3n+13))/6)))) • 3T_(n+2) −3T_(n+1) +T_n =3(4(n+2)^2 −5(n+2)+4) −3(4(n+1)^2 −5(n+1)+4) +4n^2 −5n+4 =12(n^2 +4n+4)−15(n+2)+12 −12(n^2 +2n+1)−15(n+1)+12 +4n^2 −5n+4 =12n^2 +48n+48−15n−30+12 −12n^2 −24n−12−15n−15+12 +4n^2 −5n+4 =4n^2 −11n+19
$$\:\bullet\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{T}_{{k}} \:\:: \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|}{{T}_{{n}+\mathrm{1}} −{T}_{{n}} }&\hline{{a}_{{n}} }\\{\cancel{{T}_{\mathrm{2}} }−{T}_{\mathrm{1}} }&\hline{{a}_{\mathrm{1}} }\\{\cancel{{T}_{\mathrm{3}} }−\cancel{{T}_{\mathrm{2}} }}&\hline{{a}_{\mathrm{2}} }\\{\cancel{{T}_{\mathrm{4}} }−\cancel{{T}_{\mathrm{3}} }}&\hline{{a}_{\mathrm{3}} }\\{….}&\hline{…}\\{\cancel{{T}_{{n}} }−\cancel{{T}_{{n}−\mathrm{1}} }}&\hline{{a}_{{n}−\mathrm{1}} }\\{{T}_{{n}+\mathrm{1}} −\cancel{{T}_{{n}} }}&\hline{{a}_{{n}} }\\{{T}_{{n}+\mathrm{1}} −{T}_{\mathrm{1}} }&\hline{\frac{{n}}{\mathrm{2}}\left\{\mathrm{2}{a}_{\mathrm{1}} +\left({n}−\mathrm{1}\right){d}\right\}}\\\hline\end{array}\: \\ $$$${T}_{{n}+\mathrm{1}} −{T}_{\mathrm{1}} =\frac{{n}}{\mathrm{2}}\left\{\mathrm{2}\left(\mathrm{7}\right)+\mathrm{8}\left({n}−\mathrm{1}\right)\right\} \\ $$$${T}_{{n}+\mathrm{1}} −{T}_{\mathrm{1}} ={n}\left\{\mathrm{7}+\mathrm{4}\left({n}−\mathrm{1}\right)\right\} \\ $$$${T}_{{n}+\mathrm{1}} −\mathrm{3}={n}\left(\mathrm{4}{n}+\mathrm{3}\right) \\ $$$${T}_{{n}+\mathrm{1}} =\mathrm{4}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3} \\ $$$${T}_{{n}} =\mathrm{4}\left({n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left({n}−\mathrm{1}\right)+\mathrm{3} \\ $$$$\:\:\:\:\:\:=\mathrm{4}\left({n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{3}{n}−\mathrm{3}+\mathrm{3} \\ $$$$\:\:\:\:\:=\mathrm{4}{n}^{\mathrm{2}} −\mathrm{8}{n}+\mathrm{4}+\mathrm{3}{n} \\ $$$$\:\:\:\:=\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{4} \\ $$$$\begin{array}{|c|}{{T}_{{n}} =\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{4}}\\\hline\end{array} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{T}_{{k}} =\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}{k}^{\mathrm{2}} −\mathrm{5}\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}{k}+\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{4}\left(\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\right)−\mathrm{5}\left(\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right)+\mathrm{4}{n} \\ $$$$\:\:\:=\frac{\mathrm{2}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{3}}−\frac{\mathrm{5}{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+\mathrm{4}{n} \\ $$$$=\frac{\mathrm{4}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{15}{n}\left({n}+\mathrm{1}\right)+\mathrm{24}{n}}{\mathrm{6}} \\ $$$$={n}\left(\frac{\mathrm{8}{n}^{\mathrm{2}} +\mathrm{12}{n}+\mathrm{4}−\mathrm{15}{n}−\mathrm{15}+\mathrm{24}}{\mathrm{6}}\right) \\ $$$$={n}\left(\frac{\mathrm{8}{n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{13}}{\mathrm{6}}\right) \\ $$$$=\frac{{n}\left(\mathrm{8}{n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{13}\right)}{\mathrm{6}} \\ $$$$\begin{array}{|c|}{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{T}_{{k}} =\frac{{n}\left(\mathrm{8}{n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{13}\right)}{\mathrm{6}}}\\\hline\end{array} \\ $$$$\: \\ $$$$\bullet\:\:\mathrm{3}{T}_{{n}+\mathrm{2}} −\mathrm{3}{T}_{{n}+\mathrm{1}} +{T}_{{n}} \: \\ $$$$\:\:=\mathrm{3}\left(\mathrm{4}\left({n}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{5}\left({n}+\mathrm{2}\right)+\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{3}\left(\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}\left({n}+\mathrm{1}\right)+\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{4} \\ $$$$\:\:=\mathrm{12}\left({n}^{\mathrm{2}} +\mathrm{4}{n}+\mathrm{4}\right)−\mathrm{15}\left({n}+\mathrm{2}\right)+\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:−\mathrm{12}\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{15}\left({n}+\mathrm{1}\right)+\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{4} \\ $$$$=\mathrm{12}{n}^{\mathrm{2}} +\mathrm{48}{n}+\mathrm{48}−\mathrm{15}{n}−\mathrm{30}+\mathrm{12} \\ $$$$\:\:\:\:\:−\mathrm{12}{n}^{\mathrm{2}} −\mathrm{24}{n}−\mathrm{12}−\mathrm{15}{n}−\mathrm{15}+\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{4} \\ $$$$=\mathrm{4}{n}^{\mathrm{2}} −\mathrm{11}{n}+\mathrm{19} \\ $$
Commented by Mingma last updated on 12/Feb/23
Excellent

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *