Q on recurrence with algebra

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Sonal7

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Oct 4, 2019
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I cant do the algebra.

Question has the following equation:
gn=g1+[MATH]\sum_{r=2}^{n} (r-1)[/MATH]
g1=0

So gn= [MATH]\sum_{r=2}^{n}r[/MATH]-[MATH]\sum_{r=2}^{n}1[/MATH]
part of the questions asks to show that gn=[MATH]\frac{n(n-1)}{2}[/MATH]
I cant see the answer but unsure how it works. I got as far as:
gn=[MATH]\frac{n(n+1)}{2} + n[/MATH]I understand the LHS of the solution below but I cant seem to make it work.
1584118006190.png
 
sorry it does work:

I just expanded it and it matches the RHS.

gn= [MATH]\frac{n(n+1)}{2}-1-(n-1)[/MATH]=[MATH]\frac{n(n+1)-2-2(n-1)}{2}[/MATH]=[MATH]\frac{n2-n}{2}[/MATH]=[MATH]\frac{n(n-1)}{2}[/MATH]
 
\(\sum\limits_{r = 2}^n r - \sum\limits_{r = 2}^n 1 = \left[ {\frac{{n(n + 1)}}{2} - 1} \right] - \left[ {n - 1} \right]\)
 
Sonal7 said:
I cant do the algebra.

Question has the following equation:
gn=g1+[MATH]\sum_{r=2}^{n} (r-1)[/MATH]
g1=0

So gn= [MATH]\sum_{r=2}^{n}r[/MATH]-[MATH]\sum_{r=2}^{n}1[/MATH]
part of the questions asks to show that gn=[MATH]\frac{n(n-1)}{2}[/MATH]
I cant see the answer but unsure how it works. I got as far as:
gn=[MATH]\frac{n(n+1)}{2} + n[/MATH]
Click to expand...
That is wrong. It is minus n rather than plus n.
[MATH]\text {GIVEN: } g_1 = 0 \text { and } g_n = g_1 + \left ( \sum_{r=2}^n r - 1 \right ). [/MATH]
I am presuming that the second formula says if n > 1. Otherwise nothing makes sense.

[MATH]g_1 = 0 \text { and } g_n = g_1 + \left ( \sum_{r=2}^n r - 1 \right ) \implies g_n = 0 + \left ( \sum_{r=2}^n r \right ) - \left ( \sum_{r=2}^n 1 \right ) \implies[/MATH]
[MATH]g_n =\left \{ \left ( \sum_{r=1}^n r \right ) - 1 \right \} - \left \{ \left ( \sum_{r=1}^n 1 \right ) - 1 \right \} = \dfrac{n(n + 1)}{2} - 1 - (n - 1) \implies[/MATH]
[MATH]g_n = \dfrac{n(n + 1)}{2} - n.[/MATH]
Did you make a typo?

I understand the LHS of the solution below but I cant seem to make it work.
View attachment 17161
Click to expand...
[MATH]g_n = \dfrac{n(n + 1)}{2} - n = \dfrac{n^2 + n}{2} - \dfrac{2n}{2} = \dfrac{n^2 + n - 2n}{2} \implies[/MATH]
[MATH]g_n = \dfrac{n^2 - n}{2} = \dfrac{n(n -1)}{2}.[/MATH]
 
This is so hard. No i merely stumbled my way around. Heres the question which explains why the equation is what it is. I will put the full solution to part a which sets one up to the rest of the question. This makes a lot more sense. I feel less confused :). The question is referring number of matches played between different teams and I found the problem interesting.
1584120550306.png

1584120619794.png
 
I would have done....
\(\displaystyle \sum_{r=2}^n{(r-1)} = \sum_{r=1}^{n-1}r = \dfrac{n(n-1)}{2}\)
 
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