For every integer n ≥ 3, the function g(n) is defined as the

The odd integers from 1 to 100, inclusive = the odd integers from 1 to 99, inclusive
[because 100 is even]

Therefore g(100) = g(99)
=> g(100) - g(99) = 0

(A)

Just understanding what stem is trying to convey:

for every integer \(n \ge 3\), g(n)=Product of odd integers from 1 to n:

g(3): Here, n=3
So, g(3)=1*3(Product of ODD integers from 1 to n i.e. 1 to 3)

g(4): Here, n=4
So, g(4)=1*3 (Product of ODD integers from 1 to n i.e. 1 to 4)

Likewise,
g(99): Here, n=99
So, g(99)=1*3*5*7*....*95*97*99 (Product of ODD integers from 1 to n i.e. 1 to 99)

g(100): Here, n=100
So, g(100)=1*3*5*7*....*95*97*99 (Product of ODD integers from 1 to n i.e. 1 to 100)

You see g(99) and g(100) are same. Thus the difference must be 0.

Ans: "A"
********************************

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For every integer n ≥ 3, the function g(n) is defined as the product of all the odd integers from 1 to n, inclusive. What is the value of g(100) – g(99)?

1. 0
2. 1
3. 3
4. 99
5. 100

g(n) is defined as product of all odd integers from 1 to n.

Lets consider a case :

let n =7 ;then g(n)=3*5*7 =75
let n =8 ;then g(n)=3*5*7 =75

There is no change in the final value of both the functions .This is because there is no odd number between 7 and 8 and hence the value is same.

The same is the case with 99 and 100. i.e. g(100) will be same as g(99)

=> g(100)-g(99) = 0 which is the answer choice A.

@gsaxena26: As per your request, here is my solution.

First, we will first try and understand what the function g(n) is. g(n) is product of all odd integers upto and including n. Let me take some examples to understand this.

g(3) = 1*3 (Odd integers upto and including 3)
g(4) = 1*3 (Odd integers upto and including 4)
g(5) = 1*3*5 (Odd integers upto and including 5)
g(6) = 1*3*5 (Odd integers upto and including 6)
etc

Ok. Now we have to deal with g(99) and g(100). Let's get a sense of what they are.

g(99) = 1*3*5*7*9*...*99
g(100) = 1*3*5*7*9*...*99
They are going to be the same. Hence their difference will be 0.

Takeaway: If it is hard to understand what the question is saying, try and take some examples to understand it. Here, we tried to figure out what g(n) is using g(3), g(4) etc

for n>=3 , g(n) = product of all odd integers 1 to n.

g(100) = 1*3*5.....99

g(99) = 1*3*5*.....99

=> g(100)-g(99) = 0

Answer is A.

Hey Karishma,
i got this question correct.
But i am curious to know what if the question were g(102) – g(99) ?
how would you solve the question ?

The logic remains the same:

g(102) = 1*3*5*7*9*...*99*101
g(99) = 1*3*5*7*9*...*99

g(102) - g(99) = 1*3*5*7*9*...*99*101 - 1*3*5*7*9*...*99 = 1*3*5*7*9*...*99*(101 - 1) = 1*3*5*7*9*...*99*100

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