Greatest Term of Binomial Expansion/Examples/Arbitrary Example 1

Theorem
Consider the expression:


 * $E = \paren {1 + 2 x}^{10 \frac 1 2}$

Let $x = \dfrac 3 7$.

Then the greatest term in the power series expansion of $E$ by means of the General Binomial Theorem is:


 * $\dfrac {21 \times 19 \times 17 \times 15 \times 13} {5!} \paren {\dfrac 3 7}^5$

Proof
Let us perform the expansion:
 * $\paren {1 + 2 x}^{\frac {21} 2} = 1 + \dfrac {21} 2 \paren {2 x} + \dfrac {\paren {\frac {21} 2} \paren {\frac {19} 2} } {2!} \paren {2 x}^2 + \dfrac {\paren {\frac {21} 2} \paren {\frac {19} 2} \paren {\frac {17} 2} } {3!} \paren {2 x}^3 + \cdots$

Consider the $\paren {r + 1}$th term:


 * $u_{r + 1} = \dfrac {\paren {\frac {21} 2} \paren {\frac {19} 2} \cdots \paren {\frac {23} 2 - r} } {r!} \paren {2 x}^r$

Hence $u_{r + 1} > u_r$ if:

So:
 * if $r < 5 \tfrac 4 {13}$ then $u_{r + 1} > u_r$

but conversely:
 * if $r > 5 \tfrac 4 {13}$ then $u_{r + 1} < u_r$

Hence the greatest value of $r$ for which $u_{r + 1} > u_r$ is $5$.

Then we have that:
 * $u_6 > u_5$

but then:
 * $u_7 < u_6$

Hence the $6$th term is greatest: