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Given sequence xn=1+3/2^2+5/2^4+7/2^6+--+(2n+1)/2^2n, find limit xn with help xn - (xn)/4?

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xalal | Honors

Posted May 28, 2013 at 5:19 PM via web

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Given sequence xn=1+3/2^2+5/2^4+7/2^6+--+(2n+1)/2^2n, find limit xn with help xn - (xn)/4?

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted May 28, 2013 at 7:13 PM (Answer #1)

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Find `lim_(n->oo)x_n` if `x_n=1+3/4+5/16+7/64+ ...+(2n+1)/(2^(2n))`

Consider `x_n-x_n/4` :

`x_n/4=1/4+3/16+5/64+7/256+ ...`

`x_n-x_n/4=(1+3/4+5/16+7/64+...)-(1/4+3/16+5/64+7/256+...)`

Distribute the minus sign and add like terms:

`=1+1/2+1/8+1/32+1/128+...`

Ignoring the 1, we have a geometric series with `a_n=1/2,r=1/4` so we can sum the right hand side for infinite n:

`=1+(1/2)/(1-1/4)=1+2/3=5/3`

So `x_n-x_n/4=5/3`

`==>4x_n-x_n=20/3`

`==>3x_n=20/3`

`==>x_n=20/9`

Thus the limit is `20/9`

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