Question

The sum of infinite series \(1-3x+5x^2-7x^3+....to \infty\) when |x|<1 is:
a)\[\frac{1-x}{1+x^2}\]
b)\[\frac{1+x}{1-x^2}\]
c)\[\frac{1-x}{(1+x)^2}\]
d)\[\frac{1-x}{1-x^2}\]

Answer 1

the pattern looks like
-3, 5, -7, 9, - 11 ... if you expand the denominator using binomial theorem, ,,, you might get it.

Answer 2

wait

Answer 3

http://www.wolframalpha.com/input/?i=expand+%281-x%29%2F%28x%2B1%29^2

Answer 4

I got this weird idea..
I arbitrarily put the value of x to be 0.1 and calculated up to 9x^4 (after that the number would get much smaller) which gave me a value 0.7439 and then i substituted x=0.1 in all options and BINGO!! i got 0.7438 in option C..
So, C seems to be correct!!

Answer 5

well ... that would work out too.
i was guessing
1/(1+x^2) = 1 + ..x^2 + ..x^4 + ..x^6 + ..

(1+x)(1 + ..x^2 + ..x^4 + ..x^6 + ..) = 1 + x + .a.x^2 + .a. x^3 + ... so it couldn't be 1/(x^2+ 1)

Answer 6

The answer is CCCCCCCCCCCCC

Answer 7

Woops
\[ \frac{1-x}{1-x^2} = {1 \over 1+x} = {1 \over 1 -(-x)} = 1 - x + x^2 + x^3 - ... \]

Answer 8

similarly
\[ \frac{1+x}{1-x^2} = 1 +x +x^2 + x^3 + ... \]

Answer 9

for this
\[ \frac{1-x}{1+x^2} \]
you should have alternating pattern ... which is not in the the series.
so generating function for series is C

Answer 10

Woops!! error
\[ \frac{1-x}{1-x^2} = {1 \over 1+x} = {1 \over 1 -(-x)} = 1 - x + x^2 - x^3 + ...\]

Answer 11

this is arithmetico geometric series...u just need a formula.

Answer 12

Ok here it ....
Let s =1-x+3x^2-5x^3+.....(+-)(2n-1)x^n
Now let me ask what is x^2*s

Answer 13

seems that way too ... though deriving the left from right is quite difficult.

Answer 14

And what did you get for the expression
s-x^2s

Answer 15

or @Zekarias method is also good.

Answer 16

thus s(1-x^2)=1-3x+4(x^2-x^3+x^4-.........)

Answer 17

Will you finish then?

Answer 18

Thanks everybody!!