Question

If $\underset{n\to \infty }{lim}[\left(\right({1009}^{1010}{)}^n+\left({1010}^{1009}{)}^n\right){]}^{\frac{1}{n}}$ is equal to $a^b,$ where $a,bϵN,$ then $b-a$ is equal to

Answer 1

Let $(1009{)}^{1010}=x$ and $(1010{)}^{1009}=y$
$\therefore \underset{n\to \infty }{lim}(x^n+y^n{)}^{\frac{1}{n}}=\underset{n\to \infty }{lim}x(1+(\frac{y}{x}{)}^n{)}^{\frac{1}{n}}$
As $\log x>\log y$ we have $x>y$
$\therefore \frac{y}{x}<1$
Applying binomial theorem,
$=\underset{n\to \infty }{lim}x[1+(\frac{y}{x}{)}^n.\frac{1}{n}+.....]$
$\underset{n\to \infty }{lim}(x^n+y^n{)}^{\frac{1}{n}}=\underset{n\to \infty }{lim}x=(1009{)}^{1010}=a^b$
$\therefore b-a=1$