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Find the value of
x 2 0 1 4 1 + x 2 0 1 3 1 + x 2 0 1 2 1 + . . . + x 3 1 + x 2 1 + x 1 x 2 0 1 4 + x 2 0 1 3 + x 2 0 1 2 + . . . + x 3 + x 2 + x
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3 solutions
didn't get it..... Please explain in some other way in more simplistic way please
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sorry for the inconveniece, it is a hard time for me to show simple solution because i dont know how to use latex.
i think i got a better solution :)
please go through the formula of calculating sum of a G.P. series ... u will get it
i just try this with 2^1 + 2^2 + 2^3 : 1/2^1 + 1/2^2 + 1/2^3 and the result is 2^4 ._. so the result of that question is x^2015 :v
try for x^2 :U
Apply formula for calculating sum of a G.P series.... then simplify it answer will come
First, simplify the denominator.
x 2 0 1 4 1 + x + x 2 + . . . + x 2 0 1 3 x 2 0 1 4 + x 2 0 1 3 + . . . + x 2 + x 1
factor out one x in the numerator, then reciprocate the denominator to make solution simplier
x ( x 2 0 1 3 + . . . + x 2 + x 1 + 1 ) ⋅ 1 + x + x 2 + . . . + x 2 0 1 3 x 2 0 1 4
you can now cancel the long part of the solution,
x ⋅ x 2 0 1 4 = x 2 0 1 5