if lim x rightarrow 1 x+x sequre + x qube +……..+xn – n
Question:

If $\lim _{x \rightarrow 1} \frac{x+x^{2}+x^{3}+\ldots+x^{n}-n}{x-1}=820,(n \in N)$ then

the value of $\mathrm{n}$ is equal to

Solution:

$\lim _{x \rightarrow 1} \frac{x+x^{2}+\ldots \ldots+x^{2}-n}{x-1}=820$

$\Rightarrow \lim _{x \rightarrow 1}\left(\frac{x-1}{x-1}+\frac{x^{2}-1}{x-1}+\ldots . . \frac{x^{n}-1}{x-1}\right)=820$

$\Rightarrow 1+2+\ldots . .+\mathrm{n}=820$

$\Rightarrow \mathrm{n}(\mathrm{n}+1)=2 \times 820$

$\Rightarrow \mathrm{n}(\mathrm{n}+1)=40 \times 41$

Since $n \in N$, so $n=40$

Administrator

Leave a comment

Please enter comment.
Please enter your name.