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Question Number 3974 by Yozzii last updated on 26/Dec/15

Define the sequence {a_n } by the recursive formula a_(n+1) =ca_n −nr^(n−1) (n∈Z,n≥1) with a_1 =h and c,r,h∈C, c,r≠0. Find a_n in terms of n.

$${Define}\:{the}\:{sequence}\:\left\{{a}_{{n}} \right\}\:{by}\:{the} \\ $$$${recursive}\:{formula}\: \\ $$$$\:\:\:{a}_{{n}+\mathrm{1}} ={ca}_{{n}} −{nr}^{{n}−\mathrm{1}} \:\:\:\left({n}\in\mathbb{Z},{n}\geqslant\mathrm{1}\right) \\ $$$${with}\:{a}_{\mathrm{1}} ={h}\:{and}\:{c},{r},{h}\in\mathbb{C},\:{c},{r}\neq\mathrm{0}. \\ $$$${Find}\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$

Commented by Rasheed Soomro last updated on 26/Dec/15

a_(n+1) =ca_n −nr^(n−1) (n∈Z,n≥1) n→n−1 a_n =ca_(n−1) −(n−1)r^(n−2) a_2 =ch−1 a_3 =c(ch−1)−2r=c^2 h−c−2r a_4 =c(c^2 h−c−2r)−3r^2 =c^3 h−c^2 −2cr−3r^2 a_5 =c(c^3 h−c^2 −2cr−3r^2 )−4r^3 =c^4 h−c^3 −2c^2 r−3cr^2 −4r^3 Observing/Copying pattern { ((a_1 =h)),((a_n ^(n≥2) ={−c^(n−2) −2c^(n−3) r−3c^(n−4) r^2 −...−k c^(n−k−1) r^(k−1) }+c^(n−1) h, 1≤k≤n)) :}

$$\:\:{a}_{{n}+\mathrm{1}} ={ca}_{{n}} −{nr}^{{n}−\mathrm{1}} \:\:\:\left({n}\in\mathbb{Z},{n}\geqslant\mathrm{1}\right) \\ $$$${n}\rightarrow{n}−\mathrm{1} \\ $$$${a}_{{n}} ={ca}_{{n}−\mathrm{1}} −\left({n}−\mathrm{1}\right){r}^{{n}−\mathrm{2}} \\ $$$${a}_{\mathrm{2}} ={ch}−\mathrm{1} \\ $$$${a}_{\mathrm{3}} ={c}\left({ch}−\mathrm{1}\right)−\mathrm{2}{r}={c}^{\mathrm{2}} {h}−{c}−\mathrm{2}{r} \\ $$$${a}_{\mathrm{4}} ={c}\left({c}^{\mathrm{2}} {h}−{c}−\mathrm{2}{r}\right)−\mathrm{3}{r}^{\mathrm{2}} ={c}^{\mathrm{3}} {h}−{c}^{\mathrm{2}} −\mathrm{2}{cr}−\mathrm{3}{r}^{\mathrm{2}} \\ $$$${a}_{\mathrm{5}} ={c}\left({c}^{\mathrm{3}} {h}−{c}^{\mathrm{2}} −\mathrm{2}{cr}−\mathrm{3}{r}^{\mathrm{2}} \right)−\mathrm{4}{r}^{\mathrm{3}} ={c}^{\mathrm{4}} {h}−{c}^{\mathrm{3}} −\mathrm{2}{c}^{\mathrm{2}} {r}−\mathrm{3}{cr}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{3}} \\ $$$$\mathcal{O}{bserving}/\mathcal{C}{opying}\:\:{pattern} \\ $$$$\begin{cases}{{a}_{\mathrm{1}} ={h}}\\{\overset{{n}\geqslant\mathrm{2}} {{a}_{{n}} }=\left\{−{c}^{{n}−\mathrm{2}} −\mathrm{2}{c}^{{n}−\mathrm{3}} {r}−\mathrm{3}{c}^{{n}−\mathrm{4}} {r}^{\mathrm{2}} −...−{k}\:{c}^{{n}−{k}−\mathrm{1}} {r}^{{k}−\mathrm{1}} \right\}+{c}^{{n}−\mathrm{1}} {h},\:\mathrm{1}\leqslant{k}\leqslant{n}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by prakash jain last updated on 26/Dec/15

From Rasheed′s contribution. a_n =−c^(n−2) −2c^(n−3) r−3c^(n−4) r^2 −...−k c^(n−k−1) r^(k−1) +c^(n−1) h a_n =c^(n−1) h−Σ_(k=0) ^(n−2) (n−2−k)c^k r^(n−2−k) (r/c)a_n =((c^(n−1) h)/r)−Σ_(k=0) ^(n−2) (n−2−k)c^(k−1) r^(n−1−k) b_n =c^(n−2) +2c^(n−1) r+3c^(n−2) r^2 +4c^(n−3) r^3 +..+(n−1)r^(n−2) (r/c)b_n = +c^(n−1) r+2c^(n−2) r+3c^(n−3) r^3 +..+(n−2)r^(n−2) +(n−1)(r^(n−2) /c) b_n (1−(r/c))=c^(n−2) +c^(n−1) r+c^(n−2) r^2 +...+r^(n−2) +(n−1)(r^(n−2) /c) red is a GP with Acommon ratio (r/c) term (n−1) b_n (((c−r)/c))=((c^(n−2) ((r^(n−1) /c^(n−1) )−1))/((r/c)−1))+(n−1)(r^(n−2) /c) b_n (((c−r)/c))=(((r^(n−1) −c^(n−1) ))/(r−c))+(n−1)(r^(n−2) /c) b_n =−((c(r^(n−1) −c^(n−1) ))/((c−r)^2 ))+(((n−1)r^(n−2) )/((c−r))) a_n =c^(n−1) h−b_n a_n =c^(n−1) h+((c(r^(n−1) −c^(n−1) ))/((c−r)^2 ))−(((n−1)r^(n−2) )/((c−r)))

$$\mathrm{From}\:\mathrm{Rasheed}'\mathrm{s}\:\mathrm{contribution}. \\ $$$${a}_{{n}} =−{c}^{{n}−\mathrm{2}} −\mathrm{2}{c}^{{n}−\mathrm{3}} {r}−\mathrm{3}{c}^{{n}−\mathrm{4}} {r}^{\mathrm{2}} −...−{k}\:{c}^{{n}−{k}−\mathrm{1}} {r}^{{k}−\mathrm{1}} +{c}^{{n}−\mathrm{1}} {h} \\ $$$${a}_{{n}} ={c}^{{n}−\mathrm{1}} {h}−\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\sum}}\left({n}−\mathrm{2}−{k}\right){c}^{{k}} {r}^{{n}−\mathrm{2}−{k}} \\ $$$$\frac{{r}}{{c}}{a}_{{n}} =\frac{{c}^{{n}−\mathrm{1}} {h}}{{r}}−\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\sum}}\left({n}−\mathrm{2}−{k}\right){c}^{{k}−\mathrm{1}} {r}^{{n}−\mathrm{1}−{k}} \\ $$$${b}_{{n}} ={c}^{{n}−\mathrm{2}} +\mathrm{2}{c}^{{n}−\mathrm{1}} {r}+\mathrm{3}{c}^{{n}−\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{4}{c}^{{n}−\mathrm{3}} {r}^{\mathrm{3}} +..+\left({n}−\mathrm{1}\right){r}^{{n}−\mathrm{2}} \\ $$$$\frac{{r}}{{c}}{b}_{{n}} =\:\:\:\:\:+{c}^{{n}−\mathrm{1}} {r}+\mathrm{2}{c}^{{n}−\mathrm{2}} {r}+\mathrm{3}{c}^{{n}−\mathrm{3}} {r}^{\mathrm{3}} +..+\left({n}−\mathrm{2}\right){r}^{{n}−\mathrm{2}} +\left({n}−\mathrm{1}\right)\frac{{r}^{{n}−\mathrm{2}} }{{c}} \\ $$$${b}_{{n}} \left(\mathrm{1}−\frac{{r}}{{c}}\right)={c}^{{n}−\mathrm{2}} +{c}^{{n}−\mathrm{1}} {r}+{c}^{{n}−\mathrm{2}} {r}^{\mathrm{2}} +...+{r}^{{n}−\mathrm{2}} +\left({n}−\mathrm{1}\right)\frac{{r}^{{n}−\mathrm{2}} }{{c}} \\ $$$${red}\:{is}\:{a}\:\mathrm{GP}\:\mathrm{with}\:\mathrm{Acommon}\:\mathrm{ratio}\:\frac{{r}}{{c}}\:{term}\:\left({n}−\mathrm{1}\right) \\ $$$${b}_{{n}} \left(\frac{{c}−{r}}{{c}}\right)=\frac{{c}^{{n}−\mathrm{2}} \left(\frac{{r}^{{n}−\mathrm{1}} }{{c}^{{n}−\mathrm{1}} }−\mathrm{1}\right)}{\frac{{r}}{{c}}−\mathrm{1}}+\left({n}−\mathrm{1}\right)\frac{{r}^{{n}−\mathrm{2}} }{{c}} \\ $$$${b}_{{n}} \left(\frac{{c}−{r}}{{c}}\right)=\frac{\left({r}^{{n}−\mathrm{1}} −{c}^{{n}−\mathrm{1}} \right)}{{r}−{c}}+\left({n}−\mathrm{1}\right)\frac{{r}^{{n}−\mathrm{2}} }{{c}} \\ $$$${b}_{{n}} =−\frac{{c}\left({r}^{{n}−\mathrm{1}} −{c}^{{n}−\mathrm{1}} \right)}{\left({c}−{r}\right)^{\mathrm{2}} }+\frac{\left({n}−\mathrm{1}\right){r}^{{n}−\mathrm{2}} }{\left({c}−{r}\right)} \\ $$$${a}_{{n}} ={c}^{{n}−\mathrm{1}} {h}−{b}_{{n}} \\ $$$${a}_{{n}} ={c}^{{n}−\mathrm{1}} {h}+\frac{{c}\left({r}^{{n}−\mathrm{1}} −{c}^{{n}−\mathrm{1}} \right)}{\left({c}−{r}\right)^{\mathrm{2}} }−\frac{\left({n}−\mathrm{1}\right){r}^{{n}−\mathrm{2}} }{\left({c}−{r}\right)} \\ $$

Commented by Yozzii last updated on 27/Dec/15

Awesome.

$${Awesome}. \\ $$

Commented by Rasheed Soomro last updated on 27/Dec/15

GreaT!

$$\mathbb{G}\mathfrak{rea}\mathbb{T}! \\ $$

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