Binomial theorem with inequality I was reading the resolution of a problem and came across the following:
$\dfrac{T_{p+1}}{T_p}=\dfrac{121 - p}{2p}; T_{p+1} > T_p$ if $p \leq 40$ and $T_{p+1} < T_p$ if $p \geq 41$
for $p \in \mathbb{N}$ and $p \geq 0$.
Can anyone help me clarify these inequalities?
The problem is:
Determine the maximum term in the development of $(1 + \dfrac{1}{2})^{120}$
 A: Notice that $f(p):=\frac{121-p}{2p}=1$ for $p=40+\frac{1}{3}$. You can show that $0<f(p)<1$ for $120\ge p\ge 41$ and $f(p)>1$ for $0<p\le 40$. Therefore, 


*

*If $0< p\le 40$, $1< \frac{T_{p+1}}{T_p}$. 

*If $41\le p\le 120$, $0< \frac{T_{p+1}}{T_p}<1$. 


If you multiply the two inequalities by $T_p$ you will get the desired result but you have to prove that $T_p>0$ for $0<p\le 120$ which can be done by induction. 
A: Just solve the inequality:
$$\frac{121-p}{2p}>1 \overset{p>0}{\Longrightarrow} 121-2p > p \Longrightarrow 121 > 3p \Longrightarrow  p < \frac{121}{3} \Longrightarrow p \leqslant 40$$
And same for other one.
A: Note:
this is one of my nothing-original-here
answers.
Consider the expansion
$(a+b)^n
=\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}
$.
To find the maxmum term,
consider the ratio of
consecutive terms.
$\begin{array}\\
r(k)
&=\dfrac{\binom{n}{k+1}a^{k+1}b^{n-(k+1)}}{\binom{n}{k}a^kb^{n-k}}\\
&=\dfrac{\dfrac{n!}{(k+1)!(n-k-1)!}a^{k+1}b^{n-k-1}}{\dfrac{n!}{k!(n-k)!}a^kb^{n-k}}\\
&=\dfrac{(n-k)a}{(k+1)b}\\
\end{array}
$
so
$r(k) = 1$ when
$(n-k)a
=(k+1)b
$
or
$na-ka
=kb+b
$
or
$na-b
=k(a+b)
$
or
$k
=\dfrac{na-b}{a+b}
=nu-(1-u)
=v(n, a, b)
$
where
$u = \dfrac{a}{a+b}
$.
If
$k \le v(n, a, b)$,
then
$r(k) \le 1$;
If
$k \ge v(n, a, b)$,
then
$r(k) \ge 1$.
Therefore the maximum is at
$w(n, a, b)
=\lfloor v(n, a, b) \rfloor
$
or
$w(n, a, b)+1
$.
