General Term in the Expansion of (a+b)^n

Naming the terms. In the expansion of $(a+b)^n$, the successive terms are labelled $t_1,t_2,t_3,\ldots,t_r,t_{r+1},\ldots,t_{n+1}$, so that $t_1={}^nC_0a^nb^0$, $t_2={}^nC_1a^{n-1}b^1$, $t_3={}^nC_2a^{n-2}b^2$, and in general $t_r={}^nC_{r-1}a^{n-r+1}b^{r-1}$, so that $t_{r+1}={}^nC_r\,a^{n-r}\,b^r,\qquad 0\le r\le n.$ This $t_{r+1}$ is called the general term of the expansion: because it is written in terms of $r$ rather than being tied to any one specific term, substituting the value of $r$ that matches whatever is asked for (a particular term number, or a particular power of the variable) picks that term out directly, without expanding the whole binomial.

Solved Example 1. Find the fifth term in the expansion of $\left(2x^2+\dfrac{3}{2x}\right)^8$. Solution. Here $a=2x^2$, $b=\dfrac{3}{2x}$, $n=8$. For $t_5$, $r=4$ (since $t_{r+1}=t_5\Rightarrow r=4$). $t_5={}^8C_4(2x^2)^{8-4}\left(\dfrac{3}{2x}\right)^4={}^8C_4(2x^2)^4\left(\dfrac{3}{2x}\right)^4=70(16x^8)\left(\dfrac{81}{16x^4}\right)=70(81)x^4=5670x^4$. So the fifth term is $5670x^4$.

Solved Example 2. Find the coefficient of $x^7$ in the expansion of $\left(x^2+\dfrac1x\right)^{11}$. Solution. Here $a=x^2,b=1/x,n=11$. $t_{r+1}={}^{11}C_r(x^2)^{11-r}\left(\dfrac1x\right)^r={}^{11}C_rx^{22-2r}x^{-r}={}^{11}C_rx^{22-3r}$. For the exponent of $x$ to be $7$: $22-3r=7\Rightarrow r=5$. So the coefficient is $^{11}C_5=\dfrac{11.10.9.8.7}{5.4.3.2.1}=462$.

Solved Example 3. Find the coefficient of $x^{-2}$ in the expansion of $\left(2x^2-\dfrac{1}{3x}\right)^{10}$. Solution. Here $a=2x^2,b=-\dfrac{1}{3x},n=10$. $t_{r+1}={}^{10}C_r(2x^2)^{10-r}\left(-\dfrac{1}{3x}\right)^r={}^{10}C_r(2)^{10-r}\left(-\dfrac13\right)^rx^{10-3r}$. Setting $10-3r=-2\Rightarrow r=4$. Coefficient $={}^{10}C_4(2)^6\left(-\dfrac13\right)^4=210(64)\dfrac19=\dfrac{4480}{3}$. …