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

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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Using binomial theorem we can say;


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

`=sum_(r=0)^6 ^6C_r x^r (x^-2)^(6-r)2^r`

`=sum_(r=0)^6 ^6C_r 2^r x^(3r-12)`

For the term independent of x;

`3^r-12 = 0`

`r = 4`

Since first term refers to r = 0 fifth term is given by r = 4

So the fifth term will be independent of  x