Qwertyuiop[]
Junior Member
- Joined
- Jun 1, 2022
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- 123
A sequence is defined by [imath]u_0 = 3[/imath] and [imath]u_n=\frac{u_{n-1}+2n^2-2}{n^2}[/imath] for N*.
The question is show by induction that for [imath]n \in N, u_n \ge 2.[/imath]. I want to show you my working, i just want to know if it's correct and i didn't make any mistakes in the induction.
Base case: [imath]u_0 = 3[/imath] and [imath]u_0 \ge 2[/imath] so holds for [imath]u_0[/imath].
Inductive hyp: Assuming [imath]u_k \ge 2[/imath], we have to prove that [imath]u_{k+1}\:\ge \:2[/imath].
[imath]u_{k+1}\:is\:given\:by\:u_{k+1}=\:\frac{u_k+2\left(k+1\right)^2-2}{\left(k+1\right)^2}[/imath].
making [imath]u_k[/imath] the subject we get : [imath]u_k\:=\:\left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2+2[/imath] and using the assumption [imath]u_k \ge 2[/imath] ; [imath]\left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2+2 \ge 2 \\ \left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2 \ge 0 \\ \left(k+1\right)^2\:\left(u_{k+1}\:-2\right)\:\ge 0[/imath] . Dividing by [imath]\left(k+1\right)^2[/imath] because it's positive so the sign of inequality is conserved and because we have n >1 we are not dividing by zero.
so [imath]u_{k+1}\ge 2[/imath]
if it's true for [imath]u_{k+1}[/imath] then it must be true [imath]\forall n \in N[/imath].
The question is show by induction that for [imath]n \in N, u_n \ge 2.[/imath]. I want to show you my working, i just want to know if it's correct and i didn't make any mistakes in the induction.
Base case: [imath]u_0 = 3[/imath] and [imath]u_0 \ge 2[/imath] so holds for [imath]u_0[/imath].
Inductive hyp: Assuming [imath]u_k \ge 2[/imath], we have to prove that [imath]u_{k+1}\:\ge \:2[/imath].
[imath]u_{k+1}\:is\:given\:by\:u_{k+1}=\:\frac{u_k+2\left(k+1\right)^2-2}{\left(k+1\right)^2}[/imath].
making [imath]u_k[/imath] the subject we get : [imath]u_k\:=\:\left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2+2[/imath] and using the assumption [imath]u_k \ge 2[/imath] ; [imath]\left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2+2 \ge 2 \\ \left(u_{k+1}\right)\left(k+1\right)^2-2\left(k+1\right)^2 \ge 0 \\ \left(k+1\right)^2\:\left(u_{k+1}\:-2\right)\:\ge 0[/imath] . Dividing by [imath]\left(k+1\right)^2[/imath] because it's positive so the sign of inequality is conserved and because we have n >1 we are not dividing by zero.
so [imath]u_{k+1}\ge 2[/imath]
if it's true for [imath]u_{k+1}[/imath] then it must be true [imath]\forall n \in N[/imath].