# Find the term independent of x in the expansion of `(2x+1/x^2)^6` .

*print*Print*list*Cite

### 1 Answer

This can be done using binomial expansion. Using binomial expansion we can say;

`(a+b)^n = sum_(nr=0)^n^nC_ra^rb^(n-r)`

`(2x+1/x^2)^6 = sum_(r = 0)^6^6C_r(2x)^rxx(1/x^2)^(6-r)`

`(2x+1/x^2)^6 = sum_(r = 0)^6^6C_r2^rx^(r-2(6-r))`

When we have terms independent of x then;

`r-2(6-r) = 0`

`3r = 12`

`r = 4`

So when r = 4 we have a term without x.

`T_4 = ^6C_4xx2^4`

`T_4 = 240`

*So the term without x is 240.*

**Sources:**