The number of tournament games is represented as G(n) where n is the number of attendees of the games. 2 attendees play a game such that G(n+1)=G(n)+n, G(2)=1. If the attendees number is 30, what is the total number of games?
A. 380
B. 435
C. 455
D. 510
E. 520
The OA is B.
I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
The number of tournament games is represented as...
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G(n+1)=G(n)+nLUANDATO wrote:The number of tournament games is represented as G(n) where n is the number of attendees of the games. 2 attendees play a game such that G(n+1)=G(n)+n, G(2)=1. If the attendees number is 30, what is the total number of games?
A. 380
B. 435
C. 455
D. 510
E. 520
The OA is B.
I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
G(2)=1
G(3)=G(2)+2 = 1+2 = 3
G(4)=G(3)+3 = 1+2 + 3
G(5)=G(4)+4 = 1+2 + 3+4
i.e. G(30)=1+2 + 3+4+5+.......+29 = (1/2)*29*(29+1) = 29*15 = 435
Answer: option B
I hope this helps!
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Hello LUANDATO.
In this question we will use the Gauss Sum, which says that the sum of the integers from 1 to n is $$\frac{n\cdot\left(n+1\right)}{2}.$$
It tell us $$G\left(n+1\right)=G\left(n\right)+n, and \ G\left(2\right)=1.$$
So, $$G\left(3\right)=G\left(2\right)+2=1+2.$$ $$G\left(4\right)=G\left(3\right)+3=1+2+3.$$ $$G\left(5\right)=G\left(4\right)+4=1+2+3\ +5.$$ As you can see, we can say that $$G\left(n\right)=1+2+3 +.\ .\ .\ +\ \left(n-1\right).$$ In other words, G(n) is the Gauss Sum from 1 to n-1.
So, $$G\left(30\right)=1+2+3 +\ .\ .\ .\ +\ 29\ =\ \frac{29\cdot30}{2}=15\cdot29=435.$$
The correct answer is B .
I hope this explanation may help you.
I'm available if you'd like a follow up.
In this question we will use the Gauss Sum, which says that the sum of the integers from 1 to n is $$\frac{n\cdot\left(n+1\right)}{2}.$$
It tell us $$G\left(n+1\right)=G\left(n\right)+n, and \ G\left(2\right)=1.$$
So, $$G\left(3\right)=G\left(2\right)+2=1+2.$$ $$G\left(4\right)=G\left(3\right)+3=1+2+3.$$ $$G\left(5\right)=G\left(4\right)+4=1+2+3\ +5.$$ As you can see, we can say that $$G\left(n\right)=1+2+3 +.\ .\ .\ +\ \left(n-1\right).$$ In other words, G(n) is the Gauss Sum from 1 to n-1.
So, $$G\left(30\right)=1+2+3 +\ .\ .\ .\ +\ 29\ =\ \frac{29\cdot30}{2}=15\cdot29=435.$$
The correct answer is B .
I hope this explanation may help you.
I'm available if you'd like a follow up.
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We see that G(3) = G(2) + 2 = 1 + 2 = 3, G(4) = G(3) + 3 = 3 + 3 = 6, G(5) = G(4) + 4 = 6 + 4 = 10, and so on. In other words, if n ≥ 2, G(n) = 1 + 2 + 3 + ... + (n - 1). Therefore, G(30) = 1 + 2 + 3 + ... + 29. In other words, G(30) is the sum of the first 29 positive integers. The first 29 positive integers comprise an evenly-spaced set, and so we can use the formula Sum = Average x Number to calculate the sum of the first 29 positive integers. Therefore, we have:BTGmoderatorLU wrote:The number of tournament games is represented as G(n) where n is the number of attendees of the games. 2 attendees play a game such that G(n+1)=G(n)+n, G(2)=1. If the attendees number is 30, what is the total number of games?
A. 380
B. 435
C. 455
D. 510
E. 520
The OA is B.
I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
G(30) = (1 + 29)/2 x 29 = 15 x 29 = 435
Answer: B
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