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

Σ_(k=0) ^(n−1) S_k =Σ_(k=0) ^(n−1) (k^2 +7k+2) =Σ_(k=0) ^(n−1) k^2 +7Σ_(k=0) ^(n−1) k +2n =(((n−1)(n−1+1)(2(n−1)+1))/6) +7 (((n−1)n)/2) +2n =((n(n−1)(2n−1))/6) +(7/2)n(n−1) +2n

$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{S}_{{k}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({k}^{\mathrm{2}} \:+\mathrm{7}{k}+\mathrm{2}\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}^{\mathrm{2}} \:+\mathrm{7}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\:+\mathrm{2}{n} \\ $$$$=\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{1}+\mathrm{1}\right)\left(\mathrm{2}\left({n}−\mathrm{1}\right)+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$+\mathrm{7}\:\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}\:+\mathrm{2}{n} \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{6}}\:+\frac{\mathrm{7}}{\mathrm{2}}{n}\left({n}−\mathrm{1}\right)\:+\mathrm{2}{n} \\ $$

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

(i) Un = Sn − Sn−1= n^2 +7n+2 −(n−1)^2 −7(n−1)−2 =(2n−1)+7=2n + 6

$$\left({i}\right)\:{Un}\:=\:{Sn}\:−\:{Sn}−\mathrm{1}= \\ $$$${n}^{\mathrm{2}} \:+\mathrm{7}{n}+\mathrm{2}\:−\left({n}−\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{7}\left({n}−\mathrm{1}\right)−\mathrm{2} \\ $$$$=\left(\mathrm{2}{n}−\mathrm{1}\right)+\mathrm{7}=\mathrm{2}{n}\:+\:\mathrm{6} \\ $$

Answered by Kunal12588 last updated on 30/Dec/19

T_n =S_n −S_(n−1) =n^2 −(n−1)^2 +7n−7(n−1)+2−2 =2n−1+7 =2n+6=2(n+3) ∀ n>1 as for n=1 T_1 =S_1 =10 [not S_1 −S_0 ]

$${T}_{{n}} ={S}_{{n}} −{S}_{{n}−\mathrm{1}} \\ $$$$={n}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{7}{n}−\mathrm{7}\left({n}−\mathrm{1}\right)+\mathrm{2}−\mathrm{2} \\ $$$$=\mathrm{2}{n}−\mathrm{1}+\mathrm{7} \\ $$$$=\mathrm{2}{n}+\mathrm{6}=\mathrm{2}\left({n}+\mathrm{3}\right)\:\:\forall\:{n}>\mathrm{1} \\ $$$${as}\:{for}\:{n}=\mathrm{1}\: \\ $$$${T}_{\mathrm{1}} ={S}_{\mathrm{1}} =\mathrm{10}\:\:\:\:\:\:\:\left[{not}\:{S}_{\mathrm{1}} −{S}_{\mathrm{0}} \right] \\ $$

Commented by necxxx last updated on 01/Jan/20

$${Thank}\:{you} \\ $$

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