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

1.	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» 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.

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»

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.