if the coefficient of x iin the expansion of \left(x^2+\frac{k}{x}\right)^5=270, find the value of k.
+118659 if the coefficient of x in the expansion of \(\left(x^2+\frac{k}{x}\right)^5=270, \)
find the value of k.
Using Pascal's triangle the coefficients of the binomial expansion will be 1,5,10,10,5,1
so we have
\(LHS\\ =\left(x^2+\frac{k}{x}\right)^5\\ 1*(x^2)^5*\left(\frac{k}{x}\right)^0\;\;\; +\;\;\;5*(x^2)^4*\left(\frac{k}{x}\right)^1\;\;\; +\;\;\;10*(x^2)^3*\left(\frac{k}{x}\right)^2\;\;\;\\ +\;\;\;10*(x^2)^2*\left(\frac{k}{x}\right)^3\;\;\; +\;\;\;5*(x^2)^1*\left(\frac{k}{x}\right)^4\;\;\; +\;\;\;1*(x^2)^0*\left(\frac{k}{x}\right)^5\\~\\~\\ \mbox{I am only interested in the x term}\\ 10x^4*\left(\frac{k}{x}\right)^3=270x\\ 10k^3=270\\ k^3=27\\ k=3 \)
