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)
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:
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