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
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.