Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

If
\(\lim_{x\to1}\frac{x+x^{2}+x^{3}+...x^{n}-n}{x-1} = 820\)
n∈N, then the value of n is equal to:

Solution:

Given

\(\begin{array}{l}\lim_{x\to1}\frac{(x-1)}{x-1} +\frac{(x^{2}-1)}{x-1}+..\frac{(x^{n}-1)}{x-1}= 820\end{array} \)

1+2+3+…+n = 820

Ʃn = 820

n(n+1)/2 = 820

n = 40

Answer: (40)

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