Theorem: Greatest Term of Binomial Expansion/Examples/Arbitrary Example 2

Consider the expression:
:$E = \paren {1 - 3 x}^{-\frac 7 3}$
Let $x = \dfrac 1 4$.
Then the greatest terms in the power series expansion of $E$ by means of the General Binomial Theorem are:
{{begin-eqn}}
{{eqn | o = 
      | r = \dfrac {7 \times 10 \times 13} {4^3 \times 3!}
}}
{{eqn | r = \size {-\dfrac {7 \times 10 \times 13 \times 16} {4^4 \times 4!} }
}}
{{eqn | o = \approx
      | r = 2 \cdotp 3698
      | c = 
}}
{{end-eqn}}

Proof:
Let us perform the expansion:
{{begin-eqn}}
{{eqn | l = \paren {1 - 3 x}^{-\frac 7 3}
      | r = 1 + \paren {-\dfrac 4 3 - 1} \paren {-3 x} + \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} } {2!} \paren {-3 x}^2 + \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} \paren {-\frac 4 3 - 3} } {3!} \paren {-3 x}^3 + \cdots
      | c = 
}}
{{eqn | r = 1 - \paren {-\dfrac 4 3 - 1} \paren {3 x} + \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} } {2!} \paren {3 x}^2 - \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} \paren {-\frac 4 3 - 3} } {3!} \paren {3 x}^3 + \cdots
      | c = 
}}
{{end-eqn}}
Consider the $\paren {r + 1}$th term:
:$u_{r + 1} = \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} \paren {-\frac 4 3 - 3} \cdots \paren {-\frac 4 3 - \paren {r - 1} } \paren {-\frac 4 3 - r} } {r!} \paren {-1}^r \paren {3 x}^r$
From here on in we ignore the sign, and declare the coefficient to be its absolute value.
Hence:
:$u_{r + 1} = \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} \paren {-\frac 4 3 - 3} \cdots \paren {-\frac 4 3 - \paren {r - 1} } \paren {-\frac 4 3 - r} } {r!} \paren {3 x}^r$
which leads to:
{{begin-eqn}}
{{eqn | q = 
      | l = u_{r + 1}
      | o = >
      | r = u_r
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} \paren {-\frac 4 3 - 3} \cdots \paren {-\frac 4 3 - \paren {r - 1} } \paren {-\frac 4 3 - r} } {r!} \paren {3 x}^r
      | o = >
      | r = \dfrac {\paren {-\frac 4 3 - 1} \paren {-\frac 4 3 - 2} \paren {-\frac 4 3 - 3} \cdots \paren {-\frac 4 3 - \paren {r - 1} } } {\paren {r - 1}!} \paren {3 x}^{r - 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\paren {-\frac 4 3 - r} } r \paren {3 x}
      | o = >
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\paren {\frac 4 3 + r} } r \paren {3 x}
      | o = <
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\frac 4 3 + r} \paren {\dfrac 3 4}
      | o = <
      | r = r
      | c = setting $x = \dfrac 1 4$
}}
{{eqn | ll= \leadsto
      | l = 4 + 3 r
      | o = <
      | r = 4 r
      | c = 
}}
{{eqn | ll= \leadsto
      | l = r
      | o = <
      | r = 4
      | c = 
}}
{{end-eqn}}
Hence the greatest value of $r$ for which $u_{r + 1} > u_r$ is $3$.
Then we have that:
:$u_4 > u_3$
Because this limiting value is an integer, there are two greatest values.
{{Explain|The above needs to be explored.}}
Hence the $4$th and $5$th terms are greatest:
{{begin-eqn}}
{{eqn | l = u_4
      | r = \dfrac {\paren {-\frac 7 3} \paren {-\frac {10} 3 } \paren {-\frac {13} 3}  } {3!} \paren {-3 x}^3
      | c = 
}}
{{eqn | r = \dfrac {7 \times 10 \times 13} {3^3 \times 3!} \paren {\dfrac 3 4}^3
      | c = 
}}
{{eqn | r = \dfrac {7 \times 10 \times 13} {4^3 \times 3!}
      | c = 
}}
{{eqn | o = \approx
      | r = 2 \cdotp 3698
      | c = performing the arithmetic
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = u_5
      | r = \dfrac {\paren {-\frac 7 3} \paren {-\frac {10} 3 } \paren {-\frac {13} 3} \paren {-\frac {16} 3}  } {4!} \paren {-3 x}^4
      | c = 
}}
{{eqn | r = -\dfrac {7 \times 10 \times 13 \times 16} {3^4 \times 4!} \paren {\dfrac 3 4}^4
      | c = 
}}
{{eqn | r = -\dfrac {7 \times 10 \times 13 \times 16} {4^4 \times 4!}
      | c = 
}}
{{eqn | o = \approx
      | r = 2 \cdotp 3698
      | c = performing the arithmetic
}}
{{end-eqn}}
{{qed}}