Why expand all twelve terms of $(x+2)^{11}$ when you only need the one containing $x^3$? The general term formula $T_{r+1} = \,{}^nC_r\, a^{n-r} b^r$ lets you jump straight to any term in any expansion, finding coefficients, locating constant terms, and solving "find the term containing $x^k$" in seconds. This is one of the highest-yield techniques in HSC Maths Extension 1.
In the expansion of $(x+2)^5$, you want the coefficient of $x^3$. Without expanding the full expression which value of $r$ do you think gives the term containing $x^3$? And what is the general term $T_{r+1}$? Make your best guess before reading on.
There are only two moves for every general term question. Nail both and no question in this topic can stop you:
Every general term question reduces to one of two tasks: write $T_{r+1} = \,{}^nC_r\, a^{n-r} b^r$ and then set the required power equal to the exponent in $T_{r+1}$ and solve for $r$. The off-by-one rule is the only trap: the $(r+1)$th term has index $r$, not $r+1$.
The Binomial Theorem tells us every term in $(a+b)^n$. The term that contains $b^r$, which is the $(r+1)$th term is:
Note carefully: counting starts at $r = 0$. So when $r = 0$ we get the first term $T_1 = \,{}^nC_0\, a^n$, and when $r = n$ we get the last term $T_{n+1} = \,{}^nC_n\, b^n$. This off-by-one relationship means:
To find the term containing a specific power of $x$: write out $T_{r+1}$, identify the exponent of $x$ as an expression in $r$, set it equal to the required power, and solve for $r$. If $r$ is not a whole number, the term does not exist in the expansion.
T_{r+1} = \,^nC_r\, a^{n-r} b^r, memorise this formula and its index notation; The kth term: substitute r = k-1. The off-by-one comes from counting starting at T_1, not T_0.
Pause, copy the general term formula into your book: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$; the $k$th term is found by substituting $r = k-1$; the off-by-one comes from counting starting at $T_1$, not $T_0$.
Quick check: In the expansion of $(a+b)^7$, the 4th term $T_4$ corresponds to which value of $r$?
Worked examples · 3 in a row, reveal as you go
Find the coefficient of $x^3$ in the expansion of $(x+2)^5$.
Find the 4th term in the expansion of $(2x-1)^6$.
Find the constant term in the expansion of $\left(x + \dfrac{1}{x}\right)^6$.
Did you get this? True or false: to find the 5th term of a binomial expansion, you substitute $r = 5$ into the general term formula.
Misconceptions to fix · the 3 traps that cost marks
Fill the gap: In the expansion of $(x+3)^5$, the term containing $x^4$ corresponds to $r = $ .
Activities · practice with the ideas
Find the coefficient of $x^4$ in the expansion of $(x+3)^6$. Show your working by writing the general term first.
Find the 3rd term in the expansion of $(2x+1)^5$.
Find the constant term in the expansion of $\left(x + \dfrac{1}{x}\right)^6$. Explain why this only works because 6 is even.
Find the coefficient of $x^3$ in the expansion of $(2-x)^7$.
Explain in your own words why the general term formula $T_{r+1} = \,{}^nC_r\, a^{n-r} b^r$ is more efficient than expanding the full binomial when you only need one specific term.
Earlier you were asked: which value of $r$ gives the term containing $x^3$ in $(x+2)^5$?
General term: $T_{r+1} = \,{}^5C_r \, x^{5-r} \cdot 2^r$. The exponent of $x$ is $5-r$. Setting $5-r = 3$ gives $r = 2$, so the required term is $T_3 = \,{}^5C_2 \cdot x^3 \cdot 4 = 40x^3$. The coefficient is $40$.
Why $r = 2$ and not $r = 3$? Because the formula counts how many times we chose $b$ (not $a$) from the brackets. To get $x^3 = a^3$, we chose $a$ from three brackets and $b$ from the remaining two, that's $r = 2$ copies of $b$.
Check: True or false: the constant term in $\left(x + \dfrac{1}{x}\right)^5$ is $10$.
Odd one out: Which statement about the general term $T_{r+1} = \,{}^nC_r\, a^{n-r} b^r$ is incorrect?
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the coefficient of $x^4$ in the expansion of $(x+3)^6$. Show all working. (2 marks)
Q2. Find the 3rd term in the expansion of $(2x+1)^5$. (2 marks)
Q3. (a) Find the constant term in the expansion of $\left(x + \dfrac{1}{x}\right)^6$. (b) Explain why there is no constant term in the expansion of $\left(x + \dfrac{1}{x}\right)^5$. (3 marks)
Activity: 1. $T_{r+1} = \,{}^6C_r x^{6-r} \cdot 3^r$; $x^4$: $r=2$; coefficient $= \,{}^6C_2 \cdot 3^2 = 15 \cdot 9 = 135$ · 2. $r=2$; $T_3 = \,{}^5C_2 (2x)^3 (1)^2 = 10 \times 8x^3 = 80x^3$ · 3. $r = n/2 = 3$; constant term $= \,{}^6C_3 = 20$; for odd $n$: $r = 5/2$ is not an integer so no constant term · 4. $a=2, b=-x, n=7$; $T_{r+1} = \,{}^7C_r 2^{7-r}(-x)^r = \,{}^7C_r 2^{7-r}(-1)^r x^r$; $x^3$: $r=3$; $\,{}^7C_3 \cdot 2^4 \cdot (-1)^3 = 35 \times 16 \times (-1) = -560$ · 5. The general term pinpoints the needed term with one calculation instead of generating all $n+1$ terms.
Q1 (2 marks): $T_{r+1} = \,{}^6C_r x^{6-r} \cdot 3^r$ [0.5]. For $x^4$: $6-r = 4 \Rightarrow r = 2$ [0.5]. Coefficient $= \,{}^6C_2 \times 3^2 = 15 \times 9 = 135$ [1].
Q2 (2 marks): 3rd term $\Rightarrow r = 2$ [1]. $T_3 = \,{}^5C_2 (2x)^3 \cdot 1^2 = 10 \times 8x^3 = 80x^3$ [1].
Q3 (3 marks): (a) $T_{r+1} = \,{}^6C_r x^{6-r} x^{-r} = \,{}^6C_r x^{6-2r}$ [0.5]. $6-2r = 0 \Rightarrow r = 3$ [0.5]. Constant term $= \,{}^6C_3 = 20$ [1]. (b) For $(x+1/x)^5$: exponent of $x$ is $5-2r$; setting $5-2r = 0$ gives $r = 2.5$, which is not a non-negative integer, so no constant term exists [1].
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