TITLE: Which term of the binomial expansion of $\left(1+\sqrt{2}\right)^{50}$ is the greatest?
QUESTION [5 upvotes]: Which term of the binomial expansion of $\left(1+\sqrt{2}\right)^{50}$ is the greatest?

How can I find it, without comparing all 51 values? 
Is there a quicker way to do it? (The solution says it's the number at index 30)

REPLY [4 votes]: Consider the expansion of $(1+p)^m$ which is given by
$$(1+p)^m=\sum_{r=0}^{m}\underbrace{{m\choose r  }p^r}_{u_r}$$
Note that 
$$u_{r+1}={m\choose r+1}p^{r+1}=\frac{m-r}{r+1}\color{blue}{{m\choose r}p^r}\cdot p=\left[\frac{m-r}{r+1}\cdot p\right]\color{blue}{u_r}$$
For $u_{r+1}<u_r$ (in order for $u_r$ to be the greatest term),
$$\frac{m-r}{r+1}\cdot p<1 \Rightarrow 
r>\frac{mp-1}{p+1}$$
Here $m=50, p=\sqrt2$, hence
$$r>28.8\Rightarrow r=29$$
i.e. the greatest term is the 30th term, as given by 
$$u_{29}={50\choose 29}(\sqrt2)^{29}$$
NB - the first term corresponds to $u=0$