General Binomial Theorem
\( \begin{align} \displaystyle
(a+b)^n &= \binom{n}{0}a^nb^0 + \binom{n}{1}a^{n-1}b^1 + \cdots + \binom{n}{k}a^{n-k}b^{k} + \cdots + \binom{n}{n}a^{0}b^{n} \\
&= \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}
\end{align} \)
\( \begin{align} \displaystyle
1^{\text{st}} \text{ term } T_1 &= \binom{n}{0}a^nb^0 \\
2^{\text{nd}} \text{ term } T_2 &= \binom{n}{1}a^{n-1}b^1 \\
3^{\text{rd}} \text{ term } T_3 &= \binom{n}{2}a^{n-2}b^2 \\
&\vdots \\
k^{\text{th}} \text{ term } T_k &= \binom{n}{k-1}a^{n-(k-1)}b^{k-1} \\
(k+1)^{\text{th}} \text{ term } T_{k+1} &= \binom{n}{k}a^{n-k}b^{k}
\end{align} \)
We call $\displaystyle (k+1)^{\text{th}} \text{ term } T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ as the General Term.
Example 1
Expand $(2x+3)^5$.
\( \begin{align} \displaystyle
(2x+3)^5 &= \binom{5}{0}(2x)^53^0 + \binom{5}{1}(2x)^43^1 + \binom{5}{2}(2x)^33^2 + \binom{5}{3}(2x)^23^3 + \binom{5}{4}(2x)^13^4 + \binom{5}{5}(2x)^03^5 \\
&= 2^5x^5 + 5 \times 2^4x^4 \times 3 + 10 \times 2^3x^3 \times 9 + 10 \times 2^2x^2 \times 27 + 5 \times 2x \times 81 + 243 \\
&= 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243
\end{align} \)
Example 2
Write down $5$th term of the expansion of $\displaystyle \Big(2x+\dfrac{1}{x}\Big)^{12} $.
\( \begin{align} \displaystyle
T_5 &= T_{4+1} \\
&= \binom{12}{4}(2x)^8 \Big(\dfrac{1}{x}\Big)^4 \\
&= 495 \times 256x^8 \times \dfrac{1}{x^4} \\
&= 126720x^4
\end{align} \)
Example 3
Find the coefficient of $x^6$ in the expansion of $\displaystyle\Big(x^2+\dfrac{4}{x}\Big)^{12}$.
\( \begin{align} \displaystyle
T_{k+1} &= \binom{12}{k}(x^2)^{12-k}\Big(\dfrac{4}{x}\Big)^k \\
&= \binom{12}{k}x^{24-2k}\dfrac{4^k}{x^k} \\
&= \binom{12}{k}4^k x^{24-2k-k} \\
&= \binom{12}{k}4^k x^{24-3k} \\
24-3k &= 6 \\
3k &= 18 \\
k &= 6 \\
T_{6+1} &= \binom{12}{6}4^6x^6 \\
&= 3784704
\end{align} \)
Therefore the coefficient of $x^6$ is $3784704$.
Example 4
Find the constant term in the expansion of $\displaystyle\Big(2x^3+\dfrac{1}{x}\Big)^{12}$.
\( \begin{align} \displaystyle \require{AMSsymbols} \require{color}
T_{k+1} &= \binom{12}{k}(2x^3)^{12-k}\Big(\dfrac{1}{x}\Big)^k \\
&= \binom{12}{k}2^{12-k}x^{36-3k}\Big(\dfrac{1}{x^k}\Big) \\
&= \binom{12}{k}2^{12-k}x^{36-3k-k} \\
&= \binom{12}{k}2^{12-k}x^{36-4k} \\
36-4k &= 0 &\color{red}\text{constant term}\\
k &= 9 \\
T_{9+1} &= \binom{12}{9}2^{12-9}x^0 \\
&= 1760
\end{align} \)
Therefore the constant term is $1760$.
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