Question

The value of $\frac{5050{\int }_0^1(1-x^{50}{)}^{100}dx}{{\int }_0^1(1-x^{50}{)}^{101}dx}$ is

• A. $100$
• B. $5051$
• C. $101$
• D. $5050$

Answer 1

Answer: (B)

$5051$

Let $I_n={\int }_0^1(1-x^m{)}^{n}dx$

Let $I_{n+1}={\int }_0^1(1-x^m{)}^{n+1}dx$

$={\int }_0^1(1-x^m)(1-x^m{)}^{n}dx$

$={\int }_0^1\left[\right(1-x^m{)}^n-x^m(1-x^m{)}^n]dx$

$={\int }_0^1(1-x^m{)}^{n}dx-{\int }_0^1{x}^m(1-x^m{)}^{n}dx$

$=I_n-{\int }_0^1{x}^m(1-x^m{)}^{n}dx$

$=I_n-{\int }_0^{1}x\cdot x^{m-1}(1-x^m{)}^{n}dx$

$=I_n-[-{\left|\frac{x(1-x^m{)}^{n+1}}{m(n+1)}\right|}_0^1+{\int }_0^1\frac{(1-x^m{)}^{n+1}}{m(n+1)}dx]$

$I_{n+1}=I_n-\frac{1}{m(n+1)}{\int }_0^1(1-x^m{)}^{n+1}dx$

$I_{n+1}=I_n-\frac{1}{m(n+1)}{I}_{n+1}$

$\Rightarrow I_n=I_{n+1}+\frac{1}{m(n+1)}{I}_{n+1}$

$\Rightarrow I_n=\frac{m(n+1)I_{n+1}+I_{n+1}}{m(n+1)}$

$\Rightarrow m(n+1)I_n=m(n+1)I_{n+1}+I_{n+1}...\left(1\right)$

Now $\frac{5050{\int }_0^1(1-x^{50}{)}^{100}dx}{{\int }_0^1(1-x^{50}{)}^{101}dx}=?$

Here $m=50,n=100$

Substitute in $\left(1\right)$ we get

$50\left(101\right)I_{100}=50\left(101\right)I_{101}+I_{101}$

$\Rightarrow 50\left(101\right)I_{100}=\left[50\right(101)+1]I_{101}$

$\Rightarrow \frac{I_{100}}{I_{101}}=\frac{\left[50\right(101)+1]}{50\left(101\right)}$

$\Rightarrow 5050\frac{I_{100}}{I_{101}}=5050\times \frac{\left[50\right(101)+1]}{50\left(101\right)}$

$\Rightarrow 5050\frac{I_{100}}{I_{101}}=5050\times \frac{5051}{5050}$

$\Rightarrow \frac{5050{\int }_0^1(1-x^{50}{)}^{100}dx}{{\int }_0^1(1-x^{50}{)}^{101}}=5051$