Binomial Theorem

Expanding $(a+b)^2$ or $(a+b)^3$ by hand is easy, but $(a+b)^{10}$ would mean multiplying ten brackets together. The Binomial Theorem gives the entire expansion of $(a+b)^n$ for any positive integer $n$ in one stroke, without doing the multiplication.

The numbers $\binom{n}{r}={}^{n}C_r=\dfrac{n!}{r!\,(n-r)!}$ are the binomial coefficients. Read the expansion column by column: the power of $a$ falls from $n$ down to $0$, the power of $b$ rises from $0$ up to $n$, and in every single term the two powers add to $n$. There are exactly $n+1$ terms.

Properties worth memorising: The expansion is symmetric because $\binom{n}{r}=\binom{n}{n-r}$, so coefficients read the same forwards and backwards. Putting $a=b=1$ gives $\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n$ (the sum of all coefficients), while $a=1,\,b=-1$ gives the alternating sum $\binom{n}{0}-\binom{n}{1}+\cdots=0$. A handy special case is $(1+x)^n=\sum_{r=0}^{n}\binom{n}{r}x^{r}$.

Pascal's Triangle generates these coefficients without any factorials. Each row begins and ends in $1$, and every interior number is the sum of the two numbers diagonally above it — the identity $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$, known as Pascal's rule.

Deeper Insight — why the coefficients are exactly ${}^{n}C_r$: The binomial coefficient is not a mysterious lookup number; it is a count. When you expand $(a+b)^n=(a+b)(a+b)\cdots(a+b)$ with $n$ brackets, every term in the product is formed by choosing either $a$ or $b$ from each bracket. A term $a^{n-r}b^{r}$ arises whenever you pick $b$ from exactly $r$ of the $n$ brackets and $a$ from the rest — and the number of ways to choose those $r$ brackets is precisely $\binom{n}{r}$, the number of $r$-element subsets of an $n$-element set. This is why Pascal's identity $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$ holds: a subset either contains a fixed element (choose the other $r-1$ from $n-1$) or it does not (choose all $r$ from $n-1$). The same subset-counting also explains why all the coefficients add to $2^n$: that is the total number of subsets of an $n$-set. Seeing algebra and counting as two views of one idea makes the whole chapter feel inevitable rather than memorised.

Writing out a whole expansion is wasteful when a question asks for just one term. The general term is a single formula that produces any term you want by plugging in its position.

Here $T_{r+1}$ is the $(r+1)$th term of $(a+b)^n$. The off-by-one is the only thing to watch: $r=0$ gives the first term $T_1=a^n$, $r=1$ gives the second term, and so on up to $r=n$ for the last term $T_{n+1}=b^n$. So to reach the $k$th term, set $r=k-1$. For the common form $(1+x)^n$ the general term simplifies to $T_{r+1}=\binom{n}{r}x^{r}$.

The middle term depends on whether $n$ is even or odd, because there are $n+1$ terms in all:

The logic is simple. With $n+1$ terms total, an even $n$ gives an odd count of terms, so there is one term dead in the centre at position $\frac{n}{2}+1$. An odd $n$ gives an even count, leaving two terms sharing the middle.

Deeper Insight — read $r$ as a counter, not a position: Almost every error with the general term comes from confusing the index $r$ with the term number. Think of $r$ as a counter that starts at zero — it counts how many factors of $b$ a term carries, and equivalently how many of the $n$ brackets contributed a $b$. That is exactly why $\binom{n}{r}$ appears: it counts the choices of those $r$ brackets, the same subset-counting from the previous topic. Because $r$ starts at $0$, the term number is always one more, hence $T_{r+1}$. Once you fix this in your mind, middle terms stop being a memorised rule: the centre of $n+1$ slots is at slot $\frac{n+1}{2}+\frac12=\frac{n}{2}+1$ when that is a whole number (even $n$), and otherwise the two slots straddling the centre are the middle pair. Treat the formula as a way of selecting a term and the parity question answers itself.

Many exam questions never ask for a full expansion at all. They ask for one targeted piece — the term free of $x$, the coefficient of a specific power, or the numerically greatest term. Each is solved by squeezing the right value of $r$ out of the general term.

Term independent of $x$ (the constant term): Write the general term, collect every power of $x$ into a single exponent, set that exponent equal to $0$, and solve for $r$. If that $r$ is a non-negative integer $\le n$, the constant term is $T_{r+1}$ for that $r$; if it is not an integer, no constant term exists.

Coefficient of a given power $x^m$: Same idea — set the combined exponent of $x$ equal to $m$, solve for $r$, then read off $\binom{n}{r}$ times the numerical part. Distinguish the coefficient (the number multiplying $x^m$) from the binomial coefficient $\binom{n}{r}$, since powers of constants inside the bracket also contribute.

Numerically greatest term in $(1+x)^n$ (for a fixed numerical $x$): consecutive terms are compared through their ratio

The terms increase while this ratio exceeds $1$ and decrease once it drops below $1$, so the greatest term sits where the ratio crosses $1$. Find the largest $r$ with $T_{r+1}\ge T_r$ — i.e. the integer part of $\dfrac{(n+1)|x|}{1+|x|}$ — and that locates the numerically greatest term.

Deeper Insight — every special term is the same one question: It looks like three separate techniques, but constant terms, coefficient-hunting, and greatest terms are all one move: control the exponent of $x$, then evaluate the binomial coefficient that subset-counting provides. Because $T_{r+1}=\binom{n}{r}a^{n-r}b^{r}$ packs every power into a function of the single counter $r$, you are really solving a small linear equation in $r$ to pick the term you care about. The constant term is just "power $=0$", the coefficient question is "power $=m$", and the greatest term swaps the equation for an inequality on the ratio of neighbours. Recognising this shared skeleton means you never memorise three recipes — you set up the general term once and ask what value of $r$ the question is secretly demanding. The recurring $\binom{n}{r}$ is, as always, the count of how many of the $n$ brackets supplied the second symbol.