Question

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

Answer 1

Expansion of `(3x^2+1/x)^6` has seven terms. Constant term is`135`. 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`