If the sum of the coefficient of 1st ,2nd,3rd term of the expansion of (x2+1/x) raised to the power m is 46 then find the coefficient of the terms that does not contain x?

1 Answer
Max G.
Feb 5, 2018

First find m.

Explanation:

The first three coefficients will always be
#("_0^m ) = 1#, #("_1^m ) = m#, and #("_2^m ) = (m(m-1))/2#.
The sum of these simplifies to
#m^2/2 + m/2 + 1#. Set this equal to 46, and solve for m.
#m^2/2 + m/2 + 1 = 46#
#m^2+ m + 2 = 92#
#m^2+ m - 90 = 0#
#(m + 10)(m - 9) = 0#
The only positive solution is #m = 9#.

Now, in the expansion with m = 9, the term lacking x must be the term containing #(x^2)^3(1/x)^6 = x^6/x^6 = 1#
This term has a coefficient of #("_6^9 ) = 84#.

The solution is 84.

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