Question

If the product of first $n$ odd natural numbers is $\frac{4040!}{2^{2020}\cdot 2020!}$, then the sum of these odd natural numbers is

• A. $1010\times 2020$
• B. $1010\times 3030$
• C. $1010\times 4040$
• D. $1010\times 1010$

Answer 1

The correct option is C $1010\times 4040$

Given : $1\times 3\times 5\cdots \times (2n-1)=\frac{4040!}{2^{2020}\cdot 2020!}$
We know that,
$1\times 3\times 5\cdots \times (2n-1)=\frac{2n!}{2^n\cdot n!}$
So,
$n=2020$
Now,
$1+3+5+\cdots +4039=\frac{2020}{2}[1+4039]=1010\times 4040$