IIT JEE , AIEEE Forum - Mathematics , Algebra

tejasvi2389

If an= (1000^n)/n! for n=1,2,3,..............then the sequence [ax]

a)does not have a maximum

b)attains maximum at exactly

one value of n

c)attains maximum at exactly 2 values of n

d)attains maximum at n=100

please nswer with explanation

iitkgp_bipin

1000ⁿ/n! = (1000/1).(1000/2).(1000/3)...................(1000/n)

For n = 1,2,3...........1000 this value is increasing since every term uptil here is greater than 1.

But for n>1000, 1000/n < 1 and the series would start decreasing since the series is multiplied by numbers less than 1.

Hence it attains a maximum value at n = 1000.

Best Wishes

tejasvi2389

yes I too think the option should be 1000 but CD are given as answers please help

iitkgp_bipin

At n=999 and n=1000 the values are same hence they attain maximum for two values of n.

Option d has some misprinting it should be n = 1000 instead of n = 100.

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