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Question Number 2644 by 123456 last updated on 24/Nov/15
$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{{n}}+{n} \\ $$$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{{n}} =???\:{n}\in\mathbb{N}^{\ast} \\ $$
Commented by RasheedAhmad last updated on 24/Nov/15
$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{{n}}+{n} \\ $$$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{{n}} =???\:{n}\in\mathbb{N}^{\ast} \\ $$$$−−−−−−−−−− \\ $$$${Let}\:{n}\rightarrow{n}−\mathrm{1} \\ $$$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{{n}}+{n} \\ $$$$\Rightarrow{a}_{{n}} =\frac{{a}_{{n}−\mathrm{1}} }{{n}−\mathrm{1}}+{n}−\mathrm{1} \\ $$$$ \\ $$
Answered by RasheedAhmad last updated on 24/Nov/15
![a_(n+1) =(a_n /n)+n a_1 =1 a_n =??? n∈N^∗ −−−−−−−−−− n→ n−1 a_(n+1) =(a_n /n)+n ⇒a_n =(a_(n−1) /(n−1))+n−1 a_1 =1 [given] a_2 =a_1 +1=2 a_3 =(a_2 /2)+2=3 ..... .... a_n =n This can be proved using induction.](Q2648.png)
$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{{n}}+{n} \\ $$$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{{n}} =???\:{n}\in\mathbb{N}^{\ast} \\ $$$$−−−−−−−−−− \\ $$$${n}\rightarrow\:{n}−\mathrm{1} \\ $$$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{{n}}+{n} \\ $$$$\Rightarrow{a}_{{n}} =\frac{{a}_{{n}−\mathrm{1}} }{{n}−\mathrm{1}}+{n}−\mathrm{1} \\ $$$${a}_{\mathrm{1}} =\mathrm{1}\:\:\left[{given}\right] \\ $$$$\:{a}_{\mathrm{2}} ={a}_{\mathrm{1}} +\mathrm{1}=\mathrm{2} \\ $$$${a}_{\mathrm{3}} =\frac{{a}_{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}=\mathrm{3} \\ $$$$..... \\ $$$$.... \\ $$$${a}_{{n}} ={n} \\ $$$${This}\:{can}\:{be}\:{proved}\:{using}\:{induction}. \\ $$