Q on recurrence relation

[Sonal7]

I am stuck on part b of this question. I think its a good question. I liked part a.
I got:
U₁=(2x1)-1=1
U₂=1+(2x2-1)=4
U₃=1+3+(3x2-1)=9
U₄=1+3+5+7=16

I really cant understand how they did b.
I thin they substituted Un but unclear on the rest of part b.

Ans to part b:

 

[lev888]

u_n+1 is u_n plus (2i-1) where i is n+1.

 

[Steven G]

Have you learned how to prove something via mathematical induction? Although there are easier ways to do your parts than by mathematical induction, it does work.

 

[pka]

I am stuck on part b of this question. I think its a good question. I liked part a.
I got:
U₁=(2x1)-1=1
U₂=1+(2x2-1)=4
U₃=1+3+(3x2-1)=9
U₄=1+3+5+7=16

I really cant understand how they did b.
I thin they substituted Un but unclear on the rest of part b.
View attachment 17149
Ans to part b:
View attachment 17150

Frankly I am confused by this question.
\(u_n=n^2\) to show that
\(u_n=\sum\limits_{k = 1}^n {(2k - 1)}=2\sum\limits_{k = 1}^n {k }+\sum\limits_{k = 1}^n - 1=2\dfrac{n(n+1)}{2}-n=n^2+n-n=n^2\)

 

[Sonal7]

Have you learned how to prove something via mathematical induction? Although there are easier ways to do your parts than by mathematical induction, it does work.

Yes we have, lots of similar type of questions expect you to use mathematical proof by induction. I think this isn't one of those. I needed to use n+1 for i. Thank you Lev888. I found this a hard question, however a good question. There are a few more which when solved makes one feel good.

 

[Sonal7]

Have you learned how to prove something via mathematical induction? Although there are easier ways to do your parts than by mathematical induction, it does work.

Actually I see what you mean. We are doing just that.