Question-19064

Question Number 19064 by 99 last updated on 03/Aug/17

Commented by 99 last updated on 03/Aug/17

s i r p l e a s e s o l v e t h i s q u e s t i o n

Commented by 99 last updated on 03/Aug/17

q u e s t i o n n o . 3

Answered by prakash jain last updated on 04/Aug/17

a_{n} = \underset{i = 1}{\sum^{2^{n} - 1}} \frac{1}{i}

1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + . . + \frac{1}{2^{n} - 1}

> 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + . .

+ \frac{1}{2^{n}} + \frac{1}{2^{n}} + . . + \frac{1}{2^{n}} - \frac{1}{2^{n}}

= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + {n t i m e s} - \frac{1}{2^{n}}

= 1 + \frac{n}{2} - \frac{1}{2^{n}}

a (200) > 100

- - - - - - - - - -

1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + . . + \frac{1}{2^{n} - 1}

< 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + . . + \frac{1}{2^{n - 1}} + . . + \frac{1}{2^{n - 1}}

= 1 + n - 1 = n

a (100) < 100

Commented by rahul 19 last updated on 04/Oct/18

w o w