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Question Number 150500 by EnterUsername last updated on 13/Aug/21

Find sum of first-n terms in u_n =4, 9, 15, 23, 35, 55, 91, 159,...

$$\mathrm{Find}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}-\mathrm{n}\:\mathrm{terms}\:\mathrm{in} \\ $$$${u}_{{n}} =\mathrm{4},\:\mathrm{9},\:\mathrm{15},\:\mathrm{23},\:\mathrm{35},\:\mathrm{55},\:\mathrm{91},\:\mathrm{159},... \\ $$

Commented by CAIMAN last updated on 13/Aug/21

Un=2^n-1+4n -1

Answered by Ar Brandon last updated on 13/Aug/21

u_n =4, 9, 15, 23, 35, 55, 91, 159,... Δu_n =5, 6, 8, 12, 20, 36, 68,... Δ^2 u_n =1, 2, 4, 8, 16, 32,... r=2 ⇒u_n =ar^(n−1) +a_0 +a_1 (n−1) =a2^(n−1) +a_0 +a_1 (n−1) u_1 =4=a+a_0 , u_2 =9=2a+a_0 +a_1 u_3 =15=4a+a_0 +2a_1 ⇒a=1, a_0 =3, a_1 =4 ⇒u_n =2^(n−1) +3+4(n−1)=2^(n−1) +4n−1 ⇒Σ_(k=1) ^n u_k =Σ_(k=1) ^n 2^(k−1) +4Σ_(k=1) ^n k−Σ_(k=1) ^n 1 =2^n −1+2n(n+1)−n=2^n +2n^2 +n−1

$${u}_{{n}} =\mathrm{4},\:\mathrm{9},\:\mathrm{15},\:\mathrm{23},\:\mathrm{35},\:\mathrm{55},\:\mathrm{91},\:\mathrm{159},... \\ $$$$\Delta{u}_{{n}} =\mathrm{5},\:\mathrm{6},\:\mathrm{8},\:\mathrm{12},\:\mathrm{20},\:\mathrm{36},\:\mathrm{68},... \\ $$$$\Delta^{\mathrm{2}} {u}_{{n}} =\mathrm{1},\:\mathrm{2},\:\mathrm{4},\:\mathrm{8},\:\mathrm{16},\:\mathrm{32},...\:{r}=\mathrm{2} \\ $$$$\Rightarrow{u}_{{n}} ={ar}^{{n}−\mathrm{1}} +{a}_{\mathrm{0}} +{a}_{\mathrm{1}} \left({n}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:={a}\mathrm{2}^{{n}−\mathrm{1}} +{a}_{\mathrm{0}} +{a}_{\mathrm{1}} \left({n}−\mathrm{1}\right) \\ $$$${u}_{\mathrm{1}} =\mathrm{4}={a}+{a}_{\mathrm{0}} ,\:{u}_{\mathrm{2}} =\mathrm{9}=\mathrm{2}{a}+{a}_{\mathrm{0}} +{a}_{\mathrm{1}} \\ $$$${u}_{\mathrm{3}} =\mathrm{15}=\mathrm{4}{a}+{a}_{\mathrm{0}} +\mathrm{2}{a}_{\mathrm{1}} \Rightarrow{a}=\mathrm{1},\:{a}_{\mathrm{0}} =\mathrm{3},\:{a}_{\mathrm{1}} =\mathrm{4} \\ $$$$\Rightarrow{u}_{{n}} =\mathrm{2}^{{n}−\mathrm{1}} +\mathrm{3}+\mathrm{4}\left({n}−\mathrm{1}\right)=\mathrm{2}^{{n}−\mathrm{1}} +\mathrm{4}{n}−\mathrm{1} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{u}_{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}−\mathrm{1}} +\mathrm{4}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{{n}} −\mathrm{1}+\mathrm{2}{n}\left({n}+\mathrm{1}\right)−{n}=\mathrm{2}^{{n}} +\mathrm{2}{n}^{\mathrm{2}} +{n}−\mathrm{1} \\ $$

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