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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
	$a_n =Σ_(i=1) ^(2^n −1) (1/i) 1+(1/2)+(1/3)+(1/4)+(1/5)+(1/6)+(1/7)+(1/8)+..+(1/(2^n −1)) >1+(1/2)+(1/4)+(1/4)+(1/8)+(1/8)+(1/8)+(1/8)+.. +(1/2^n )+(1/2^n )+..+(1/2^n )−(1/2^n ) =1+(1/2)+(1/2)+(1/2)+{n times}−(1/2^n ) =1+(n/2)−(1/2^n ) a(200)>100 −−−−−−−−−− 1+(1/2)+(1/3)+(1/4)+(1/5)+(1/6)+(1/7)+(1/8)+..+(1/(2^n −1)) <1+(1/2)+(1/2)+(1/4)+(1/4)+(1/4)+(1/4)+..+(1/2^(n−1) )+..+(1/2^(n−1) ) =1+n−1=n a(100)<100$
	$a_{n} = \underset{i = 1}{\sum^{2^{n} - 1}} \frac{1}{i}$ $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + . . + \frac{1}{2^{n} - 1}$ $> 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + . .$ $+ \frac{1}{2^{n}} + \frac{1}{2^{n}} + . . + \frac{1}{2^{n}} - \frac{1}{2^{n}}$ $= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + {n t i m e s} - \frac{1}{2^{n}}$ $= 1 + \frac{n}{2} - \frac{1}{2^{n}}$ $a (200) > 100$ $- - - - - - - - - -$ $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + . . + \frac{1}{2^{n} - 1}$ $< 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + . . + \frac{1}{2^{n - 1}} + . . + \frac{1}{2^{n - 1}}$ $= 1 + n - 1 = n$ $a (100) < 100$
	Commented by rahul 19 last updated on 04/Oct/18

	$w o w$
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