Question

(X^k+1/X)^6 ;at "K" is positive integer ;find value of"K' such that the expansion has a term free of X

Answer 1

@mukushla i need the value of k that make the expansion has term free of X

Answer 2

if u want to find all values of \(k\) just simplify the \[(x^k+\frac{1}{x})^6\]using binomial theorem

Answer 3

can u do that?

Answer 4

of course but at Tr+1=6Cr+(1/x)^r +(x^K)^6-r;then Tr+1=6cr (X^6k-kr-r) at term free of "X" then ; 6k-kr-r=0 and i can not complete can you help

Answer 5

very well

now u need to solve \(6k-kr-r=0\) for \(r \in \text{{0,1,2,3,4,5,6}}\)

Answer 6

because if \(6k-kr-r=0\) then exponent of \(x\) will be zero and that term

becomes \(x\) free

Answer 7

\[r \in \left\{ 0,1,2,3,4,5,6 \right\}\] becouse "r"lessthan or equal "n" and N=6 is that right?

Answer 8

exactly

Answer 9

thanks

Answer 10

yw :)