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Question Number 176662 by Matica last updated on 24/Sep/22

sequence V_(n+1) −V_n =n+3^n . Find V_n .

$$\:\:{sequence}\:{V}_{{n}+\mathrm{1}} −{V}_{{n}} ={n}+\mathrm{3}^{{n}} .\:{Find}\:{V}_{{n}} . \\ $$

Answered by mr W last updated on 24/Sep/22

V_n −V_(n−1) =n−1+3^(n−1) V_(n−1) −V_(n−2) =n−2+3^(n−2) ... V_2 −V_1 =1+3^1 Σ: V_n −V_1 =1+2+...+(n−1)+3^1 +3^2 +...+3^(n−1) V_n −V_1 =(((n−1)n)/2)+((3(3^(n−1) −1))/(3−1)) ⇒V_n =V_1 +(((n−1)n+3^n −3)/2)

$${V}_{{n}} −{V}_{{n}−\mathrm{1}} ={n}−\mathrm{1}+\mathrm{3}^{{n}−\mathrm{1}} \\ $$$${V}_{{n}−\mathrm{1}} −{V}_{{n}−\mathrm{2}} ={n}−\mathrm{2}+\mathrm{3}^{{n}−\mathrm{2}} \\ $$$$... \\ $$$${V}_{\mathrm{2}} −{V}_{\mathrm{1}} =\mathrm{1}+\mathrm{3}^{\mathrm{1}} \\ $$$$\Sigma: \\ $$$${V}_{{n}} −{V}_{\mathrm{1}} =\mathrm{1}+\mathrm{2}+...+\left({n}−\mathrm{1}\right)+\mathrm{3}^{\mathrm{1}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{3}^{{n}−\mathrm{1}} \\ $$$${V}_{{n}} −{V}_{\mathrm{1}} =\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}+\frac{\mathrm{3}\left(\mathrm{3}^{{n}−\mathrm{1}} −\mathrm{1}\right)}{\mathrm{3}−\mathrm{1}} \\ $$$$\Rightarrow{V}_{{n}} ={V}_{\mathrm{1}} +\frac{\left({n}−\mathrm{1}\right){n}+\mathrm{3}^{{n}} −\mathrm{3}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 25/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mnjuly1970 last updated on 24/Sep/22

−−−−−−−− Σ_(k=1) ^n (V_(k+1) −V_k ) = ((n(n+1))/2) +((3(1−3^( n) ))/(−2)) l.h.s(telescopic series):: V_( n+1) −V_1 = ((n(1+n))/2) +(3^( n+1) /2) −(3/2) n→ n−1 V_n = V_1 +(((n−1)n)/2) +(3^( n) /2) −(3/2) −−−−−−−−−

$$\: \\ $$$$\:\:\:\:\:\:\:\:\:−−−−−−−− \\ $$$$\:\:\:\:\:\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({V}_{{k}+\mathrm{1}} \:−{V}_{{k}} \right)\:=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:+\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{3}^{\:{n}} \right)}{−\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{l}.{h}.{s}\left({telescopic}\:{series}\right):: \\ $$$$\:\:\:\:\:\:\:\:\:\:{V}_{\:{n}+\mathrm{1}} \:−{V}_{\mathrm{1}} =\:\frac{{n}\left(\mathrm{1}+{n}\right)}{\mathrm{2}}\:+\frac{\mathrm{3}^{\:{n}+\mathrm{1}} }{\mathrm{2}}\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:{n}\rightarrow\:{n}−\mathrm{1}\:\:\:\:\:\:\:\:\:{V}_{{n}} \:=\:{V}_{\mathrm{1}} \:+\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}\:+\frac{\mathrm{3}^{\:{n}} }{\mathrm{2}}\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−−−−−− \\ $$

Answered by Mathspace last updated on 25/Sep/22

Σ_(k=0) ^n v_(k+1) −v_k =Σ_(k=0) ^n k+Σ_(k=0) ^n 3^k =((n(n+1))/2)+((1−3^(n+1) )/(1−3)) =((n(n+1))/2)−((1−3^(n+1) )/2) ⇒ v_(n+1) −v_0 =((n^2 +n−1+3^(n+1) )/2) ⇒ v_n =v_0 +(((n−1)^2 +(n−1)−1+3^n )/2) =v_0 +((n^2 −2n+1+n−2+3^n )/2) =v_0 +((3^n +n^2 −n−1)/2) with v_1 −v_0 =1

$$\sum_{{k}=\mathrm{0}} ^{{n}} {v}_{{k}+\mathrm{1}} −{v}_{{k}} =\sum_{{k}=\mathrm{0}} ^{{n}} {k}+\sum_{{k}=\mathrm{0}} ^{{n}} \mathrm{3}^{{k}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+\frac{\mathrm{1}−\mathrm{3}^{{n}+\mathrm{1}} }{\mathrm{1}−\mathrm{3}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{1}−\mathrm{3}^{{n}+\mathrm{1}} }{\mathrm{2}}\:\Rightarrow \\ $$$${v}_{{n}+\mathrm{1}} −{v}_{\mathrm{0}} =\frac{{n}^{\mathrm{2}} +{n}−\mathrm{1}+\mathrm{3}^{{n}+\mathrm{1}} }{\mathrm{2}}\:\Rightarrow \\ $$$${v}_{{n}} ={v}_{\mathrm{0}} +\frac{\left({n}−\mathrm{1}\right)^{\mathrm{2}} +\left({n}−\mathrm{1}\right)−\mathrm{1}+\mathrm{3}^{{n}} }{\mathrm{2}} \\ $$$$={v}_{\mathrm{0}} +\frac{{n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{1}+{n}−\mathrm{2}+\mathrm{3}^{{n}} }{\mathrm{2}} \\ $$$$={v}_{\mathrm{0}} +\frac{\mathrm{3}^{{n}} +{n}^{\mathrm{2}} −{n}−\mathrm{1}}{\mathrm{2}}\:{with} \\ $$$${v}_{\mathrm{1}} −{v}_{\mathrm{0}} =\mathrm{1} \\ $$

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