Consider the expansion (3x^2 + (1/x))^6 How many terms does the expansion include? Find the constant term Show that the expansion has no terms involving x^5

1 Answer
Jan 29, 2018

Expansion of (3x^2+1/x)^6 has seven terms. Constant term is135. For details see below.

Explanation:

An expansionof (a+b)^6 has seven terms and (3x^2+1/x)^6 too has seven terms. This may be seen from the following.

The expansion of (a+b)^n is C_0^na^n+C_1^na^(n-1)b^1+C_2^na^(n-2)b^2+.....+C_r^na^(n-r)b^r+...+C_n^nb^n

Observe that (r+1)^(th) term is C_r^na^(n-r)b^r

hence (r+1)^(th) term of (3x^2+1/x)^6 is

C_r^6(3x^2)^(6-r)(1/x)^r=C_r^6 3^(6-r)x^(12-3r)

Observe that r can take values only from 0 to 6, for which power of x can be 12,9,6,3,0,-3,-6,

Hence expansion does not have a term x^5.

Constant term means x^0, which is for r=4 and it is

C_4^6 3^2=(6*5*4*3)/(1*2*3*4)*9=135