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Question

$$\displaystyle \lim_{x\rightarrow 1}\frac {x+x^2+...+x^n-n}{x-1}$$ equals


A
n
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B
0
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C
n22
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D
n(n+1)2
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Solution

The correct option is D $$\displaystyle \frac {n(n+1)}{2}$$
Using L-hospital's rule,
$$\displaystyle \lim_{x\rightarrow 1}\frac {x+x^2+...+x^n-n}{x-1}$$
$$=\displaystyle \lim_{x\rightarrow 1}\frac {1+2x+3x^2+...+nx^{n-1}}{1}$$
$$=1+2+3+.....................+n=\cfrac{n}{2}(2+(n-1).1)$$
$$=\cfrac{n(n+1)}{2}$$
Hence, option 'D' is correct.

Mathematics

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