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Find the term independent of x in the expansion of `(2x+1/x^2)^6` .
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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.
Posted by jeew-m on June 9, 2013 at 2:02 AM (Answer #1)
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