Question

$\underset{x\to \infty }{lim}\frac{{(x+1)}^{100}+{(x+2)}^{100}+..+{(x+50)}^{100}}{x^{100}+x^{99}+..+x+1}=$ _____

• A. $100$
• B. $1$
• C. $21$
• D. $50$

Answer 1

The correct option is D $50$

${lim}_{x\to \infty }\frac{{(x+1)}^{100}+{(x+2)}^{100}+...+{(x+50)}^{100}}{x^{100}+x^{99}+...+x+1}$
$={lim}_{x\to \infty }\frac{x^{100}[{(1+\frac{1}{x})}^{100}+{(1+\frac{2}{x})}^{100}+...+{(1+\frac{50}{x})}^{100}]}{x^{100}[1+\frac{1}{x}+\frac{1}{x^2}+...+\frac{1}{x^{99}}+\frac{1}{x^{100}}]}$
$\frac{1+1+\mathrm{1.....50}\text{terms}}{1}[\because \frac{1}{\infty }=0]$
$=50$