The coefficient of the 5th term in the binomial expansion of \((1 + kx)^{8}\), in ascending powers of x is \(\frac{35}{8}\). Find the value of the constant ...

In the binomial expansion of \((1+kx)^8\), the coefficient of the 5th term in ascending powers of x is given by the expression: \[\binom{8}{4}(kx)^4(1)^{8-4}\] which simplifies to: \[\binom{8}{4}k^4x^4\] Using the formula for binomial coefficients, we can write: \[\binom{8}{4} = \frac{8!}{4!4!} = 70\] So, we can write the expression for the 5th term coefficient as: \[70k^4x^4\] We are given that this coefficient is \(\frac{35}{8}\), so we can write: \[70k^4x^4 = \frac{35}{8}\] Simplifying this equation, we get: \[k^4x^4 = \frac{1}{16}\] Dividing both sides by \(x^4\), we get: \[k^4 = \frac{1}{16}\] Taking the fourth root of both sides, we get: \[k = \pm \frac{1}{2}\] Since the expansion is of the form \((1+kx)^8\), the sign of k does not affect the coefficient of the 5th term, so we can take \(k = \frac{1}{2}\). Therefore, the value of the constant k is \(\frac{1}{2}\).