Question

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

Answer 1

Hint: In this question it is the given series that is \[k\left( {{n}^{2}} \right)\left( {{n}^{2}}-{{1}^{2}} \right)\left( {{n}^{2}}-{{2}^{2}} \right)\left( {{n}^{2}}-{{3}^{2}} \right).......\left[ {{n}^{2}}-{{\left( n-1 \right)}^{2}} \right]=r!\] and we have to expand the series by using properties such as \[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)\]. 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
\[k\left( {{n}^{2}} \right)\left( {{n}^{2}}-{{1}^{2}} \right)\left( {{n}^{2}}-{{2}^{2}} \right)\left( {{n}^{2}}-{{3}^{2}} \right).......\left[ {{n}^{2}}-{{\left( n-1 \right)}^{2}} \right]=r!\]
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 of\[{{a}^{2}}-{{b}^{2}}\]
So, first of all we have to expand the series by using the property of \[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)\]
After expanding this series we get:
\[k\left( {{n}^{2}} \right)\left( n-1 \right)\left( n+1 \right)\left( n-2 \right)\left( n+2 \right)\left( n-3 \right)\left( n+3 \right).......\left( n-\left( n-1 \right) \right)\left( n+\left( n-1 \right) \right)=r!\]
After simplifying this series we get:
\[k\left( {{n}^{2}} \right)\left( n-1 \right)\left( n+1 \right)\left( n-2 \right)\left( n+2 \right)\left( n-3 \right)\left( n+3 \right).......\left( 2n-1 \right)=r!\]
So in this series we can split \[{{n}^{2}}\] as \[n.n\] we get:
\[k\left( n \right)\left( n \right)\left( n-1 \right)\left( n+1 \right)\left( n-2 \right)\left( n+2 \right)\left( n-3 \right)\left( n+3 \right).......\left( 2n-1 \right)=r!\]
After solving this above series we get:
\[k.n.1.2.............\left( n-1 \right)n\left( n+1 \right)\left( n+2 \right).......\left( 2n-1 \right)=r!\]
If you observe carefully in the above series then it is look similar like a factorial
That means it is a factorial of \[\left( 2n-1 \right)\] which is represented as \[\left( 2n-1 \right)!\]
So the above series can also be written as
\[kn\left( 2n-1 \right)!=r!----(1)\]
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 \[k=2\]. Then we get the positive integer value of r
That means
Substitute \[k=2\] and \[n=1\] in equation (1) we get:
\[r!=2!\]
\[\therefore r=2\]
Hence, we get after substituting the value of \[r=2\] in equation (1).
\[2n!=2!\]
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.