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Sum-to-the-n-terms-of-the-series-whose-n-th-term-is-2-n-1-8n-3-6n-2-

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Question Number 52079 by jojosehrawat21@gmail.com last updated on 03/Jan/19
Sum to the n terms of the series whose n^(th ) term is 2^(n−1 ) + 8n^3 −6n^2
$${Sum}\:{to}\:{the}\:{n}\:{terms}\:{of}\:{the}\:{series}\:{whose}\:{n}^{{th}\:} \:{term}\:{is}\:\mathrm{2}^{{n}−\mathrm{1}\:} \:+\:\mathrm{8}{n}^{\mathrm{3}} \:−\mathrm{6}{n}^{\mathrm{2}} \\ $$
Commented by maxmathsup by imad last updated on 03/Jan/19
let U_n =2^(n−1) +8n^3 −6n^2 with U_1 =3 ⇒ Σ_(k=1) ^n U_k =Σ_(k=1) ^n 2^(k−1) +8Σ_(k=1) ^n k^3 −6 Σ_(k=1) ^n k^2 =Σ_(k=0) ^(n−1) 2^k +8((n^2 (n+1)^2 )/4) −6 ((n(n+1)(2n+1))/6) =((1−2^n )/(1−2)) +2n^2 (n+1)^2 −n(n+1)(2n+1) =2^n −1 +n(n+1){2n(n+1)−2n−1} =2^n −1 +(n^2 +n){2n^2 −1} =2^n −1 +2n^4 −n^2 +2n^3 −n Σ_(k=1) ^n U_k =2^n +2n^4 +2n^3 −n^2 −n−1 .
$${let}\:{U}_{{n}} =\mathrm{2}^{{n}−\mathrm{1}} \:+\mathrm{8}{n}^{\mathrm{3}} −\mathrm{6}{n}^{\mathrm{2}} \:\:{with}\:{U}_{\mathrm{1}} =\mathrm{3}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{U}_{{k}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\mathrm{2}^{{k}−\mathrm{1}} \:+\mathrm{8}\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{3}} \:−\mathrm{6}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{2}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\mathrm{2}^{{k}} \:+\mathrm{8}\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:−\mathrm{6}\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$=\frac{\mathrm{1}−\mathrm{2}^{{n}} }{\mathrm{1}−\mathrm{2}}\:+\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \:−{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$=\mathrm{2}^{{n}} −\mathrm{1}\:+{n}\left({n}+\mathrm{1}\right)\left\{\mathrm{2}{n}\left({n}+\mathrm{1}\right)−\mathrm{2}{n}−\mathrm{1}\right\} \\ $$$$=\mathrm{2}^{{n}} −\mathrm{1}\:+\left({n}^{\mathrm{2}} +{n}\right)\left\{\mathrm{2}{n}^{\mathrm{2}} −\mathrm{1}\right\} \\ $$$$=\mathrm{2}^{{n}} −\mathrm{1}\:+\mathrm{2}{n}^{\mathrm{4}} −{n}^{\mathrm{2}} \:+\mathrm{2}{n}^{\mathrm{3}} −{n} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{U}_{{k}} =\mathrm{2}^{{n}} \:+\mathrm{2}{n}^{\mathrm{4}} \:+\mathrm{2}{n}^{\mathrm{3}} \:−{n}^{\mathrm{2}} −{n}−\mathrm{1}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jan/19
T_n =2^(n−1) +8n^3 −6n^2 S=S_1 +S_2 −S_3 S_1 =((a(r^n −1))/(r−1))=((1(2^n −1))/(2−1))=2^n −1 S_2 =8×[((n(n+1))/2)]^2 =2n^2 (n+1)^2 S_3 =6×((n(n+1)(2n+1))/6)=n(n+1)(2n+1) S=2^(n−1) +n(n+1)(2n^2 +2n−2n−1) S=2^(n−1) +n(n+1)(2n^2 −1)
$${T}_{{n}} =\mathrm{2}^{{n}−\mathrm{1}} +\mathrm{8}{n}^{\mathrm{3}} −\mathrm{6}{n}^{\mathrm{2}} \\ $$$${S}={S}_{\mathrm{1}} +{S}_{\mathrm{2}} −{S}_{\mathrm{3}} \\ $$$${S}_{\mathrm{1}} =\frac{{a}\left({r}^{{n}} −\mathrm{1}\right)}{{r}−\mathrm{1}}=\frac{\mathrm{1}\left(\mathrm{2}^{{n}} −\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}=\mathrm{2}^{{n}} −\mathrm{1} \\ $$$${S}_{\mathrm{2}} =\mathrm{8}×\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} =\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${S}_{\mathrm{3}} =\mathrm{6}×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}={n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$${S}=\mathrm{2}^{{n}−\mathrm{1}} +{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{2}{n}−\mathrm{1}\right) \\ $$$${S}=\mathrm{2}^{{n}−\mathrm{1}} +{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}^{\mathrm{2}} −\mathrm{1}\right) \\ $$

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