QOD-24/05/13 summation of series

BITSAT 2015 Preparation › Question Of the Day questions › QOD-24/05/13 summation of series

Find the summation
s=x+x²+x⁴+x⁸.......till infinity

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6 Answers

337

Sayan Sinha ·2013-05-24 06:38:26

s=x+x²+x⁴+x⁸.......âˆž
=>s= x(1+x²+x⁴+x⁸...âˆž)
=>s=x(1+s)
=>s=x1-x

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Shaswata Roy Check again.You made a mistake.
Upvote·0· Reply ·2013-05-24 11:01:31
Sayan Sinha In the second step, it should have been s=x(1+x+x^2+x^4...inf)
Upvote·0· Reply ·2013-05-24 23:29:55
Shaswata Roy You Sure...It's still wrong.
Upvote·0· Reply ·2013-05-25 03:01:23
Sayan Sinha I know the it should have been x^3..x^5 and so on. So, how about taking x^2 common?
Upvote·0· Reply ·2013-05-28 02:40:26
Sayan Sinha Nope. Doesn't work...
Upvote·0· Reply ·2013-05-28 02:45:51
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2305

Shaswata Roy ·2013-05-24 11:11:11

I think s = x²-x2x-1

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Shaswata Roy I guess it is wrong.
Upvote·0· Reply ·2013-05-24 23:29:24
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466

Himanshu Giria ·2013-05-25 06:49:15

IF X<1,
THEN S=1+X;
IF Xâ‰¥1 ,
THEN S=âˆž

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Asish Mahapatra A simple check : at x=0, S = 0 :)
Upvote·0· Reply ·2013-05-26 04:01:43
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466

Himanshu Giria ·2013-05-25 06:51:13

IF X<0 ,
THEN S=1-X;

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Himanshu Giria NO SORRY IT WILL B 1+X ONLY
Upvote·0· Reply ·2013-05-25 06:52:02
Sayan Sinha My solution works that way, right??? :)
Upvote·0· Reply ·2013-05-26 23:43:52
Sayantan Hazra Simple check : At x = -1, acc. to u, S = 0. But is it?
Upvote·0· Reply ·2013-05-27 12:08:46
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206

Sayantan Hazra ·2013-05-27 12:09:55

Hint: Concentrate on 2nd,3rd,4th terms leaving aside the 1st term.... :)

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Shaswata Roy ·2013-05-27 23:48:11

This is a Lacunary function [See-"http://en.wikipedia.org/wiki/Lacunary_function"].

It does not have a closed form.

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