Question

Evaluate: $\underset{x\to \infty }{lim}\frac{(x+1{)}^{10}+(x+2{)}^{10}+\dots ..+(x+100{)}^{10}}{x^{10}+{10}^{10}}$

• A. $10$
• B. $100$
• C. $1000$
• D. $1$

Answer 1

The correct option is B $100$

Given: $\underset{x\to \infty }{lim}\frac{(x+1{)}^{10}+(x+2{)}^{10}+\dots ..+(x+100{)}^{10}}{x^{10}+{10}^{10}}$

$\Rightarrow \underset{x\to \infty }{lim}\frac{(x+1{)}^{10}+(x+2{)}^{10}+\dots ..+(x+100{)}^{10}}{x^{10}+{10}^{10}}$
$=\underset{x\to \infty }{lim}\frac{x^{10}\left(\right(1+\frac{1}{x}{)}^{10}+(1+\frac{2}{x}{)}^{10}.....+(1+\frac{100}{x}{)}^{10})}{x^{10}(1+\frac{{10}^{10}}{x^{10}})}$
$=\underset{x\to \infty }{lim}\frac{\left(\right(1+\frac{1}{x}{)}^{10}+(1+\frac{2}{x}{)}^{10}.....+(1+\frac{100}{x}{)}^{10})}{(1+\frac{{10}^{10}}{x^{10}})}$

$=\frac{1+0+1+0+....1+0\left(100\text{terms}\right)}{1+0}$ $[\because n\to \infty \Rightarrow \frac{1}{n}\to 0]$

$=\frac{1+1+....+\left(100\text{terms}\right)}{1}$
$=100$

Hence option ${}^{\prime }{B}^{\prime }$ is the answer.