# a-n-1-a-n-n-n-a-1-1-a-n-n-N-

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
$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{{n}}+{n} \\$$$${a}_{\mathrm{1}} =\mathrm{1} \\$$$${a}_{{n}} =???\:{n}\in\mathbb{N}^{\ast} \\$$$$−−−−−−−−−− \\$$$${Let}\:{n}\rightarrow{n}−\mathrm{1} \\$$$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{{n}}+{n} \\$$$$\Rightarrow{a}_{{n}} =\frac{{a}_{{n}−\mathrm{1}} }{{n}−\mathrm{1}}+{n}−\mathrm{1} \\$$$$\\$$
$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{{n}}+{n} \\$$$${a}_{\mathrm{1}} =\mathrm{1} \\$$$${a}_{{n}} =???\:{n}\in\mathbb{N}^{\ast} \\$$$$−−−−−−−−−− \\$$$${n}\rightarrow\:{n}−\mathrm{1} \\$$$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{{n}}+{n} \\$$$$\Rightarrow{a}_{{n}} =\frac{{a}_{{n}−\mathrm{1}} }{{n}−\mathrm{1}}+{n}−\mathrm{1} \\$$$${a}_{\mathrm{1}} =\mathrm{1}\:\:\left[{given}\right] \\$$$$\:{a}_{\mathrm{2}} ={a}_{\mathrm{1}} +\mathrm{1}=\mathrm{2} \\$$$${a}_{\mathrm{3}} =\frac{{a}_{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}=\mathrm{3} \\$$$$….. \\$$$$…. \\$$$${a}_{{n}} ={n} \\$$$${This}\:{can}\:{be}\:{proved}\:{using}\:{induction}. \\$$