Greatest Term of Binomial Expansion/Examples/Arbitrary Example 1

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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:

\(\ds u_{r + 1}\) \(>\) \(\ds u_r\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\paren {\frac {21} 2} \paren {\frac {19} 2} \cdots \paren {\frac {23} 2 - r} } {r!} \paren {2 x}^r\) \(>\) \(\ds \dfrac {\paren {\frac {21} 2} \paren {\frac {19} 2} \cdots \paren {\frac {25} 2 - r} } {\paren {r - 1}!} \paren {2 x}^{r - 1}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\paren {\frac {23} 2 - r} } r \paren {2 x}\) \(>\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\paren {\frac {23} 2 - r} } r \paren {\dfrac 6 7}\) \(>\) \(\ds r\) setting $x = \dfrac 3 7$
\(\ds \leadsto \ \ \) \(\ds 69 - 6 r\) \(>\) \(\ds 7 r\)
\(\ds \leadsto \ \ \) \(\ds 13 r\) \(<\) \(\ds 69\)
\(\ds \leadsto \ \ \) \(\ds r\) \(<\) \(\ds 5 \tfrac 4 {13}\)

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:

\(\ds u_6\) \(=\) \(\ds \dfrac {\paren {\frac {21} 2} \paren {\frac {19} 2} \paren {\frac {17} 2} \paren {\frac {15} 2} \paren {\frac {13} 2} } {5!} \paren {2 x}^5\)
\(\ds \) \(=\) \(\ds \dfrac {21 \times 19 \times 17 \times 15 \times 13} {2^5 \times 5!} \paren {\dfrac 6 7}^5\)
\(\ds \) \(=\) \(\ds \dfrac {21 \times 19 \times 17 \times 15 \times 13} {5!} \paren {\dfrac 3 7}^5\)

$\blacksquare$


Sources

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