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Question: The least positive integer K for which\[k\left( {{n}^{2}} \right)\left( {{n}^{2}}-{{1}^{2}} \right)\...

The least positive integer K for whichk(n2)(n2−12)(n2−22)(n2−32).......[n2−(n−1)2]=r!k\left( {{n}^{2}} \right)\left( {{n}^{2}}-{{1}^{2}} \right)\left( {{n}^{2}}-{{2}^{2}} \right)\left( {{n}^{2}}-{{3}^{2}} \right).......\left[ {{n}^{2}}-{{\left( n-1 \right)}^{2}} \right]=r! for some positive integers r is?
A) 2002
B) 2004
C) 1
D) 2

Explanation

Solution

In this question it is the given series that is k(n2)(n2−12)(n2−22)(n2−32).......[n2−(n−1)2]=r!k\left( {{n}^{2}} \right)\left( {{n}^{2}}-{{1}^{2}} \right)\left( {{n}^{2}}-{{2}^{2}} \right)\left( {{n}^{2}}-{{3}^{2}} \right).......\left[ {{n}^{2}}-{{\left( n-1 \right)}^{2}} \right]=r! and we have to expand the series by using properties such as a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=(a-b)(a+b). Now, we have to select the value of least positive integer in such a way that the value of r will be positive integer. So, this is the approach for such types of problems.

Complete step by step answer:
According to the given series is that
k(n2)(n2−12)(n2−22)(n2−32).......[n2−(n−1)2]=r!k\left( {{n}^{2}} \right)\left( {{n}^{2}}-{{1}^{2}} \right)\left( {{n}^{2}}-{{2}^{2}} \right)\left( {{n}^{2}}-{{3}^{2}} \right).......\left[ {{n}^{2}}-{{\left( n-1 \right)}^{2}} \right]=r!
We have to find the values of k such that that positive integer of r
If you observe carefully in series it is in the form ofa2−b2{{a}^{2}}-{{b}^{2}}
So, first of all we have to expand the series by using the property of a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=(a-b)(a+b)
After expanding this series we get:
k(n2)(n−1)(n+1)(n−2)(n+2)(n−3)(n+3).......(n−(n−1))(n+(n−1))=r!k\left( {{n}^{2}} \right)\left( n-1 \right)\left( n+1 \right)\left( n-2 \right)\left( n+2 \right)\left( n-3 \right)\left( n+3 \right).......\left( n-\left( n-1 \right) \right)\left( n+\left( n-1 \right) \right)=r!
After simplifying this series we get:
k(n2)(n−1)(n+1)(n−2)(n+2)(n−3)(n+3).......(2n−1)=r!k\left( {{n}^{2}} \right)\left( n-1 \right)\left( n+1 \right)\left( n-2 \right)\left( n+2 \right)\left( n-3 \right)\left( n+3 \right).......\left( 2n-1 \right)=r!
So in this series we can split n2{{n}^{2}} as n.nn.n we get:
k(n)(n)(n−1)(n+1)(n−2)(n+2)(n−3)(n+3).......(2n−1)=r!k\left( n \right)\left( n \right)\left( n-1 \right)\left( n+1 \right)\left( n-2 \right)\left( n+2 \right)\left( n-3 \right)\left( n+3 \right).......\left( 2n-1 \right)=r!
After solving this above series we get:
k.n.1.2.............(n−1)n(n+1)(n+2).......(2n−1)=r!k.n.1.2.............\left( n-1 \right)n\left( n+1 \right)\left( n+2 \right).......\left( 2n-1 \right)=r!
If you observe carefully in the above series then it is look similar like a factorial
That means it is a factorial of (2n−1)\left( 2n-1 \right) which is represented as (2n−1)!\left( 2n-1 \right)!
So the above series can also be written as
kn(2n−1)!=r!−−−−(1)kn\left( 2n-1 \right)!=r!----(1)
So we have to select least integer K in such a way that we get the positive integer of r
If we substitute the value of k=2k=2. Then we get the positive integer value of r
That means
Substitute k=2k=2 and n=1n=1 in equation (1) we get:
r!=2!r!=2!
∴r=2\therefore r=2
Hence, we get after substituting the value of r=2r=2 in equation (1).
2n!=2!2n!=2!
Therefore, option (D) is correct.

Note:
Always remember in this type of problem is that we have to expand the series by using the correct property of basic property. Don’t make a mistake while selecting the value of K because we can’t take the value which gives a negative integer. According to the question we have to select the least integer values of K which gives a positive integer value of r. See the question what is asked and the condition which is given then solve the problems.