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

Answered by MM42 last updated on 10/Jul/23

$$u_n=u_{n-1}+n+1;n⩾2,u_1=4$$$$u_n=u_{n-1}+(n+1)$$$$=u_{n-2}+\left(n\right)+(n+1)$$$$=u_{n-3}+(n-1)+\left(n\right)+(n+1)$$$$⋮$$$$=u_2+4+5+...+\left(n\right)+(n+1)$$$$=u_1+3+4+...+\left(n\right)+(n+1)$$$$\Rightarrow u_n=4+\frac{(n-1)(n+4)}{2}=\frac{n^2}{2}+\frac{3n}{2}+2✓$$

Answered by Frix last updated on 10/Jul/23

$$4+3=7$$$$7+4=11$$$$11+5=16$$$$u_n=\frac{n^2}{2}+\frac{3n}{2}+2$$$$u_{100000}=5000150002$$

Commented by BaliramKumar last updated on 10/Jul/23

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

Commented by Frix last updated on 10/Jul/23

$$u_n=c_2{n}^2+c_{1}n+c_0$$$$\mathrm{Insert}\mathrm{all}\mathrm{given}\mathrm{pairs}(n,u_n)\mathrm{and}\mathrm{solve}\mathrm{the}$$$$\mathrm{system}.$$

Answered by horsebrand11 last updated on 11/Jul/23

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

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