The general term and finding a required term

Unit 22 · Binomial Expansion

Write the general term $T_{r+1} = {}^nC_r\,a^{n-r}b^r$, find the $k$th term, and find the coefficient of a given power of $x$ by solving for $r$.

The general term

In $(a+b)^n = \sum_{r=0}^{n} {}^nC_r, a^{n-r} b^r$, the term with $b^r$ is called the general term: $$T_{r+1} = {}^nC_r,a^{n-r},b^{r}.$$ The subscript is $r+1$ because $r$ starts at $0$: $T_1$ (first term, $r=0$) is $a^n$, $T_2$ (second term, $r=1$) is ${}^nC_1a^{n-1}b$, and so on. So the $k$th term is $T_k$ with $r = k - 1$.

Example: Find the 5th term of $(2x+3)^7$. The 5th term is $T_5$, so $r = 4$, $a = 2x$, $b = 3$, $n = 7$. $T_5 = {}^7C_4(2x)^{3}(3)^{4} = 35\cdot8x^3\cdot81 = 22680,x^3$.

Example: Write the 3rd term of $\left(1 - \dfrac{x}{2}\right)^{10}$. $r = 2$, $a = 1$, $b = -\dfrac{x}{2}$: $T_3 = {}^{10}C_2(1)^8\left(-\dfrac x2\right)^2 = 45\cdot\dfrac{x^2}{4} = \dfrac{45}{4}x^2$.

Notice that the "term" includes the power of $x$, while its coefficient is only the number in front of it.

Finding the coefficient of a given power

To find the coefficient of $x^m$ in $(a+b)^n$ where $a$ and $b$ contain powers of $x$:

Write $T_{r+1} = {}^nC_r,a^{n-r}b^r$ with $a$ and $b$ in brackets.
Collect the powers of $x$. Using $(x^p)^q = x^{pq}$ and $x^p\cdot x^q = x^{p+q}$ get $x^{\text{(something in } r)}$.
Equate the exponent to $m$ and solve for $r$. Check $r$ is an integer between $0$ and $n$.
Substitute this $r$ into the numerical part: ${}^nC_r$ times the numbers and signs from $a$ and $b$.

Example: Find the coefficient of $x^4$ in $\left(x^2 - \dfrac{2}{x}\right)^8$. $T_{r+1} = {}^8C_r(x^2)^{8-r}\left(-\dfrac2x\right)^r = {}^8C_r(-2)^r,x^{16-2r-r} = {}^8C_r(-2)^r,x^{16-3r}$. For $x^4$: $16 - 3r = 4$, so $r = 4$. Coefficient $= {}^8C_4(-2)^4 = 70\cdot16 = 1120$.

Example: Find the coefficient of $x^{-3}$ in $\left(2x - \dfrac{1}{x^2}\right)^9$. $T_{r+1} = {}^9C_r(2x)^{9-r}\left(-x^{-2}\right)^r = {}^9C_r,2^{9-r}(-1)^r,x^{9-r-2r} = {}^9C_r,2^{9-r}(-1)^r,x^{9-3r}$. For $x^{-3}$: $9 - 3r = -3$, so $r = 4$. Coefficient $= {}^9C_4\cdot2^5\cdot(+1) = 126\cdot32 = 4032$.

When no integer $r$ with $0 \le r \le n$ gives the required exponent, that power does not occur and its coefficient is $0$. For instance, in $\left(x^2 + \dfrac1x\right)^{6}$ the exponents are $12 - 3r$, all multiples of $3$, so the coefficient of $x^5$ is $0$.

Unknown constants are found the same way: express the coefficient in terms of the unknown, then use the given value.

Exam tip: විභාග පරීක්ෂකවරු "$T_{r+1} = \dots$" ලිවීමට, නිවැරදි $\binom nr$ ට, $r$ සඳහා ඇති දර්ශක සමීකරණයට සහ $r$ හි අගයට ලකුණු දෙති. මෙම පේළි තුන ම ලියන්න. $r$ සොයාගත් පසු, ඉල්ලූ $x$ හි බලය ලැබෙන දැයි ආදේශ කර නැවත පරීක්ෂා කරන්න. $x^{-1}$ යනු $-1$ බලය බව මතක තබාගන්න: $x$ හි දර්ශකය සහ පදයේ අංකය $r$ පටලවා නොගන්න.