Question Number 219476 by universe last updated on 26/Apr/25
$$\:\mathrm{let}\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{1}\:;\:\left(\mathrm{n}+\mathrm{1}\right)\mathrm{a}_{\mathrm{n}+\mathrm{1}} +\mathrm{na}_{\mathrm{n}} \:=\:\mathrm{2n}−\mathrm{3}\: \\ $$$$\:\:\mathrm{find}\:\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{a}_{\mathrm{n}\:} \\ $$
Commented by universe last updated on 26/Apr/25
$${yes}\:{sir}\: \\ $$
Commented by SdC355 last updated on 26/Apr/25
$${A}_{{n}} =\frac{\left(−\right)^{{n}} \left(−{c}_{\mathrm{1}} +\left({n}−\mathrm{2}\right)\left(−\right)^{{n}} +\mathrm{2}\right)}{{n}} \\ $$$${A}_{\mathrm{1}} =−\left(−\mathrm{c}_{\mathrm{1}} +\mathrm{3}\right)=\mathrm{1} \\ $$$$\mathrm{c}_{\mathrm{1}} −\mathrm{3}=\mathrm{1}\:\therefore\mathrm{c}_{\mathrm{1}} =\mathrm{4} \\ $$$${A}_{{n}} =\frac{\left(−\right)^{{n}} \left(\left({n}−\mathrm{2}\right)\left(−\right)^{{n}} −\mathrm{2}\right)}{{n}} \\ $$
Answered by mr W last updated on 26/Apr/25
![(n+1)a_(n+1) +na_n =2n−3 (n+1)(a_(n+1) −1)+2+n(a_n −1)+2=0 let b_n =n(a_n −1)+2 b_(n+1) +b_n =0 b_(n+1) =−b_n b_n =−b_(n−1) =...=(−1)^(n−1) b_1 =(−1)^(n−1) [1×(a_1 −1)+2]=2(−1)^(n−1) a_n =((b_n −2)/n)+1=((2(−1)^(n−1) −2)/n)+1 =1−((2[1+(−1)^n ])/n) ⇒a_n = { ((1−(4/n) for n=even)),((1 for n=odd)) :}](https://www.tinkutara.com/question/Q219480.png)
$$\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{1}} +{na}_{{n}} =\mathrm{2}{n}−\mathrm{3} \\ $$$$\left({n}+\mathrm{1}\right)\left({a}_{{n}+\mathrm{1}} −\mathrm{1}\right)+\mathrm{2}+{n}\left({a}_{{n}} −\mathrm{1}\right)+\mathrm{2}=\mathrm{0} \\ $$$${let}\:{b}_{{n}} ={n}\left({a}_{{n}} −\mathrm{1}\right)+\mathrm{2} \\ $$$${b}_{{n}+\mathrm{1}} +{b}_{{n}} =\mathrm{0} \\ $$$${b}_{{n}+\mathrm{1}} =−{b}_{{n}} \\ $$$${b}_{{n}} =−{b}_{{n}−\mathrm{1}} =…=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {b}_{\mathrm{1}} \\ $$$$\:\:\:\:=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left[\mathrm{1}×\left({a}_{\mathrm{1}} −\mathrm{1}\right)+\mathrm{2}\right]=\mathrm{2}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \\ $$$${a}_{{n}} =\frac{{b}_{{n}} −\mathrm{2}}{{n}}+\mathrm{1}=\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} −\mathrm{2}}{{n}}+\mathrm{1} \\ $$$$\:\:\:\:=\mathrm{1}−\frac{\mathrm{2}\left[\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right]}{{n}} \\ $$$$\Rightarrow{a}_{{n}} =\begin{cases}{\mathrm{1}−\frac{\mathrm{4}}{{n}}\:{for}\:{n}={even}}\\{\mathrm{1}\:{for}\:{n}={odd}}\end{cases} \\ $$