$\mathop {\lim }\limits_{x \to 1} \frac{{\mathop \sum \nolimits_{r = 1}^n \,\,{x^r} - n}}{{x - 1}}$ is equal to

JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

\lim\limits_{x \to 1} \frac{{\sum \nolimits_{r = 1}^n \,\,{x^r} - n}}{{x - 1}}

is equal to

View answer and explanation

Correct option: B

\frac{{n\left( {n + 1} \right)}}{2}

Option B (

\frac{{n\left( {n + 1} \right)}}{2}

) is correct because the limit can be evaluated by recognizing it as the sum of the first n integers.

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