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Question

Find the sum of the series 1+22x+32x2+42x3+...|x|<1.

A
(1x)(1x)3
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B
(1+x)(1x)3
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C
(1+x)(1+x)3
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D
None of these
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Solution

The correct option is C (1+x)(1x)3
Let S=1+4x+9x2+16x3+... ...(1)
Now multiplying by x throughout in (1); we get
xS=1+4x2+9x3+16x4+... ...(2)
Subtracting equation (2) from (1); we get
(1x)S=1+4x+9x2+16x3+....x4x29x316x4...
(1x)S=1+3x+5x2+7x3+.... ...(3)
Again, multiplying by x throughout in (3); we get
x(1x)S=x+3x2+5x3+7x4+.... ...(4)
Subtracting equation (4) from (3); we get
(1x)Sx(1x)S=1+x+2x+2x2+2x3+....
Notice that the series 2x+2x2+2x3+... is
geometric series with the first term a=2x and the common ratio
r=x.
Now, use the formula for the sum of an infinite geometric series.
(1x)Sx(1x)S=1+2x(1x), for|x|<1
(1x)(1x)S=1+x(1x), for|x|<1
S=1+x(1x)3, for|x|<1

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