Question-194593

Question Number 194593 by pascal889 last updated on 10/Jul/23

Answered by MM42 last updated on 10/Jul/23

u_n =u_(n−1) +n+1 ;n≥2 , u_1 =4 u_n =u_(n−1) +(n+1) =u_(n−2) +(n)+(n+1) =u_(n−3) +(n−1)+(n)+(n+1) ⋮ =u_2 +4+5+...+(n)+(n+1) =u_1 +3+4+...+(n)+(n+1) ⇒u_n =4+(((n−1)(n+4))/2)=(n^2 /2)+((3n)/2)+2 ✓

$${u}_{{n}} ={u}_{{n}−\mathrm{1}} +{n}+\mathrm{1}\:\:\:;{n}\geqslant\mathrm{2}\:\:,\:\:\:{u}_{\mathrm{1}} =\mathrm{4} \\ $$$${u}_{{n}} ={u}_{{n}−\mathrm{1}} +\left({n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:={u}_{{n}−\mathrm{2}} +\left({n}\right)+\left({n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:={u}_{{n}−\mathrm{3}} +\left({n}−\mathrm{1}\right)+\left({n}\right)+\left({n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\vdots \\ $$$$\:\:\:\:\:\:={u}_{\mathrm{2}} +\mathrm{4}+\mathrm{5}+…+\left({n}\right)+\left({n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:={u}_{\mathrm{1}} +\mathrm{3}+\mathrm{4}+…+\left({n}\right)+\left({n}+\mathrm{1}\right) \\ $$$$\Rightarrow{u}_{{n}} =\mathrm{4}+\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{4}\right)}{\mathrm{2}}=\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{3}{n}}{\mathrm{2}}+\mathrm{2}\:\:\checkmark \\ $$

Answered by Frix last updated on 10/Jul/23

4+3=7 7+4=11 11+5=16 u_n =(n^2 /2)+((3n)/2)+2 u_(100000) =5000150002

$$\mathrm{4}+\mathrm{3}=\mathrm{7} \\ $$$$\mathrm{7}+\mathrm{4}=\mathrm{11} \\ $$$$\mathrm{11}+\mathrm{5}=\mathrm{16} \\ $$$${u}_{{n}} =\frac{{n}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{3}{n}}{\mathrm{2}}+\mathrm{2} \\ $$$${u}_{\mathrm{100000}} =\mathrm{5000150002} \\ $$

Commented by BaliramKumar last updated on 10/Jul/23

please detail how to calculate u_n = (n^2 /2)+ ((3n)/2)+2

$$\mathrm{please}\:\mathrm{detail}\: \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{u}_{\mathrm{n}} \:=\:\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{2}}+\:\frac{\mathrm{3n}}{\mathrm{2}}+\mathrm{2} \\ $$

Commented by Frix last updated on 10/Jul/23

u_n =c_2 n^2 +c_1 n+c_0 Insert all given pairs (n, u_n ) and solve the system.

$${u}_{{n}} ={c}_{\mathrm{2}} {n}^{\mathrm{2}} +{c}_{\mathrm{1}} {n}+{c}_{\mathrm{0}} \\ $$$$\mathrm{Insert}\:\mathrm{all}\:\mathrm{given}\:\mathrm{pairs}\:\left({n},\:{u}_{{n}} \right)\:\mathrm{and}\:\mathrm{solve}\:\mathrm{the} \\ $$$$\mathrm{system}. \\ $$

Answered by horsebrand11 last updated on 11/Jul/23

U_n =4+S_(n−1) U_n =4+((n−1)/2)(6+(n−2).1) U_n =4+((n−1)/2)(n+4) U_n =(1/2)(n^2 +3n+4)

$$\:\mathrm{U}_{\mathrm{n}} =\mathrm{4}+\mathrm{S}_{\mathrm{n}−\mathrm{1}} \\ $$$$\:\:\mathrm{U}_{\mathrm{n}} =\mathrm{4}+\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\left(\mathrm{6}+\left(\mathrm{n}−\mathrm{2}\right).\mathrm{1}\right) \\ $$$$\:\:\mathrm{U}_{\mathrm{n}} =\mathrm{4}+\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\left(\mathrm{n}+\mathrm{4}\right) \\ $$$$\:\:\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{3n}+\mathrm{4}\right) \\ $$$$ \\ $$

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