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Question

If (1x)(1+x+x2+x3+x4) is 3132 and x is a rational number. Find the value of 1+x+x2+x3+x4+x5.

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Solution

Given (1x)(1+x+x2+x3+x4)=3132......(i)
Since, 1+x+x2+x3+x4=1x51x......(ii)
Substitute equation (ii) in eqaution (i), we get
(1x)(1x51x)=3132 or x5=13132=132
or x=1532=12
1+x+x2+x3+x4+x5=1x61x=1126112=116412
=2×6364=6332.

OR

(1x)(1+x+x2+x3+x4)=3132

1+x+x2+x3+x4xx2x3x4x5=3132

1x5=3132

x5=13132

x5=132=125

x=12

Thus,

1+x+x2+x3+x4+x5=1+12+(12)2+(12)3+(12)4+(12)5

=1+12+14+18+116+132

=6332


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