# The-sum-of-the-first-n-terms-of-a-series-is-given-by-S-n-n-2-7n-2-i-Find-a-formula-for-the-nth-term-ii-write-down-the-first-5-terms-of-the-sequence-

Question Number 76793 by necxxx last updated on 30/Dec/19
$${The}\:{sum}\:{of}\:{the}\:{first}\:{n}\:{terms}\:{of}\:{a}\:{series} \\$$$${is}\:{given}\:{by}:\:{S}_{{n}} ={n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{2}. \\$$$$\left({i}\right){Find}\:{a}\:{formula}\:{for}\:{the}\:{nth}\:{term} \\$$$$\left({ii}\right){write}\:{down}\:{the}\:{first}\:\mathrm{5}\:{terms}\:{of}\:{the} \\$$$${sequence} \\$$$$\\$$
Commented by turbo msup by abdo last updated on 30/Dec/19
$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{S}_{{k}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({k}^{\mathrm{2}} \:+\mathrm{7}{k}+\mathrm{2}\right) \\$$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}^{\mathrm{2}} \:+\mathrm{7}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\:+\mathrm{2}{n} \\$$$$=\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{1}+\mathrm{1}\right)\left(\mathrm{2}\left({n}−\mathrm{1}\right)+\mathrm{1}\right)}{\mathrm{6}} \\$$$$+\mathrm{7}\:\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}\:+\mathrm{2}{n} \\$$$$=\frac{{n}\left({n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{6}}\:+\frac{\mathrm{7}}{\mathrm{2}}{n}\left({n}−\mathrm{1}\right)\:+\mathrm{2}{n} \\$$
Commented by necxxx last updated on 01/Jan/20
$${thank}\:{you}\:{so}\:{so}\:{much}.{It}'{s}\:{clear}\:{now} \\$$
Answered by benjo 1/2 santuyy last updated on 30/Dec/19
$$\left({i}\right)\:{Un}\:=\:{Sn}\:−\:{Sn}−\mathrm{1}= \\$$$${n}^{\mathrm{2}} \:+\mathrm{7}{n}+\mathrm{2}\:−\left({n}−\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{7}\left({n}−\mathrm{1}\right)−\mathrm{2} \\$$$$=\left(\mathrm{2}{n}−\mathrm{1}\right)+\mathrm{7}=\mathrm{2}{n}\:+\:\mathrm{6} \\$$
Answered by Kunal12588 last updated on 30/Dec/19
$${T}_{{n}} ={S}_{{n}} −{S}_{{n}−\mathrm{1}} \\$$$$={n}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{7}{n}−\mathrm{7}\left({n}−\mathrm{1}\right)+\mathrm{2}−\mathrm{2} \\$$$$=\mathrm{2}{n}−\mathrm{1}+\mathrm{7} \\$$$$=\mathrm{2}{n}+\mathrm{6}=\mathrm{2}\left({n}+\mathrm{3}\right)\:\:\forall\:{n}>\mathrm{1} \\$$$${as}\:{for}\:{n}=\mathrm{1}\: \\$$$${T}_{\mathrm{1}} ={S}_{\mathrm{1}} =\mathrm{10}\:\:\:\:\:\:\:\left[{not}\:{S}_{\mathrm{1}} −{S}_{\mathrm{0}} \right] \\$$
Commented by necxxx last updated on 01/Jan/20
$${Thank}\:{you} \\$$