If n = 1000 !, then the value of (1/log2 n) + (1/log3 n)+ .... + (1/log1000 n) is

Question

If n = 1000 !, then the value of \(\frac{1}{log_2\,n}\) + \(\frac{1}{log_3\,n}\) + .... + \(\frac{1}{log_{1000}\,n}\) is

(a) 0 

(b) 1 

(c) 10 

(d) 103

Answer 1

Given, 1000! = n. Now, \(\frac{1}{log_2\,n}\) + \(\frac{1}{log_3\,n}\) + .... + \(\frac{1}{log_{1000}\,n}\) 

= log_n 2 + log_n 3 + .... + log_n 1000 \(\bigg[\text{Using}\frac{1}{log_b\,a}= log_a\,b\bigg]\)

= log_n (2 × 3 × 4 × ... × 1000) = log_n (1000!) = log_n n = 1.