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 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#