Question

Let \(x>0\) and \(n>1\) be real numbers. Prove that \((1+x)^{n}>1+n x\).

Short answer

By using the Binomial theorem, we have established that \((1+x)^{n}>1+n x\), given that \(x > 0\) and \(n > 1\), proving the given inequality is true.

Step-by-step solution

Understanding the binomial theorem

The Binomial Theorem tells us how to expand expressions of the form \((a+b)^{n}\) for any integers \(n\). According to it, we have \((1+x)^{n} = \sum_{k=0}^{n} {n\choose{k}} x^{k}\), where \({n\choose{k}}\) are binomial coefficients. When \(k=0\), this term will reduce to 1, when \(k=1\), it results in \(n\).

Apply the binomial theorem

By applying the binomial theorem, we can expand the left side of the given expression as \((1+x)^{n} = 1+ nx + \sum_{k=2}^{n} {n\choose{k}} x^{k}\). Since, each term in the series \(\sum_{k=2}^{n} {n\choose{k}} x^{k}\) is positive as \(x > 0\) and \(n > 1\), we can say that \((1+x)^{n} > 1+ nx\).

Finalize the proof

Since each term in the series \(\sum_{k=2}^{n} {n\choose{k}} x^{k}\) is positive, it must be the case that \((1+x)^{n}\) is greater than \(1+ nx\), exactly what we wanted to prove.