Question Number 35036 by abdo mathsup 649 cc last updated on 14/May/18
$${cslculate}\:\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{2}} \left({n}+\mathrm{1}−{k}\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/May/18
$${T}_{{k}} =\left({n}+\mathrm{1}\right){k}^{\mathrm{2}} −{k}^{\mathrm{3}} \\ $$$${T}_{\mathrm{1}} =\left({n}+\mathrm{1}\right)\mathrm{1}^{\mathrm{2}} −\mathrm{1}^{\mathrm{3}} \\ $$$${T}_{\mathrm{2}} =\left({n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{2}} −\mathrm{2}^{\mathrm{3}} \\ $$$${T}_{\mathrm{3}} =\left({n}+\mathrm{1}\right)\mathrm{3}^{\mathrm{2}} −\mathrm{3}^{\mathrm{3}} \\ $$$$....................... \\ $$$$........................ \\ $$$${T}_{{n}} =\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} −{n}^{\mathrm{3}} \\ $$$${so}\:{T}_{\mathrm{1}} +{T}_{\mathrm{2}} +...+{T}_{{n}} \\ $$$$=\left({n}+\mathrm{1}\right)\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−\left\{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right\}^{\mathrm{2}} \\ $$
Commented by math khazana by abdo last updated on 15/May/18
$${correct}\:{answer}\:{thanks}... \\ $$