The Binomial Theorem: Practice

Each row starts and ends with $1$, and every entry in between is the sum of the two entries directly above it.

Add the neighbouring pairs of $1,\ 4,\ 6,\ 4,\ 1$:

$1 + 4 = 5, \quad 4 + 6 = 10, \quad 6 + 4 = 10, \quad 4 + 1 = 5$

Putting a $1$ on each end gives $1,\ 5,\ 10,\ 10,\ 5,\ 1$. The new row has one more entry than the old one, which is why a six-entry answer is the right shape.

The coefficients of $(a + b)^n$ are row $n$ of the triangle, so you want row $6$.

Two checks pin it down. Row $6$ must have $6 + 1 = 7$ entries, and its second entry must be $\binom{6}{1} = 6$.

$1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1$

That row has seven entries and begins $1, 6$. The row $1,\ 5,\ 10,\ 10,\ 5,\ 1$ belongs to $(a+b)^5$, and $1,\ 7,\ 21,\ 35,\ 35,\ 21,\ 7,\ 1$ belongs to $(a+b)^7$.

Row $3$ of Pascal's triangle is $1,\ 3,\ 3,\ 1$, and the exponent on $x$ falls from $3$ to $0$ while the exponent on $y$ climbs from $0$ to $3$.

$(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$

The $3$ in front of $x^2y$ counts the three factors of $(x+y)(x+y)(x+y)$ that could have supplied the single $y$. Every term has exponents summing to $3$, which rules out $3x^2y^2$.

$\binom{7}{0}$ counts the ways for no factor to donate a $y$, which means every one of the seven factors donates its $x$.

There is exactly one way to do that, so the count is $1$, not $0$.

$\binom{n}{0} = 1 \quad \text{for every } n$

This is why every row of Pascal's triangle starts with a $1$, and it matches the fact that $(x + y)^7$ opens with a single $x^7$.

Each of the six factors donates exactly one letter, so the two exponents in any term must add up to $6$.

If two factors donated a $y$, the other four donated an $x$:

$6 - 2 = 4$

The term is $\binom{6}{2}x^4y^2 = 15x^4y^2$.

The term $x^3y^2$ takes a $y$ from $2$ of the $5$ factors, so $k = 2$ and the coefficient is $\binom{5}{2}$.

Row $5$ of Pascal's triangle is $1,\ 5,\ 10,\ 10,\ 5,\ 1$, and the entry at $k = 2$ (counting from $k = 0$) is $10$.

$\binom{5}{2} = 10$

Write the difference as a sum, $x + (-y)$, and apply the theorem with row $3$, which is $1,\ 3,\ 3,\ 1$:

$(x - y)^3 = x^3 + 3x^2(-y) + 3x(-y)^2 + (-y)^3$

Now simplify the powers of $-y$, remembering that $(-y)^2 = y^2$ is positive while $(-y)^3 = -y^3$ is negative:

$(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$

The signs alternate, and the coefficients $1, 3, 3, 1$ are unchanged.

For a difference, the coefficient of $x^{n-k}y^k$ is $(-1)^k\binom{n}{k}$, because each of the $k$ factors that donates a $y$ also donates a minus sign.

Here $k = 3$, which is odd, so the sign is negative. Row $5$ gives $\binom{5}{3} = 10$.

$(-1)^3\binom{5}{3} = -10$

The size of the coefficient is unchanged from $(x + y)^5$; only the sign flipped.

$\binom{6}{1}$ counts the ways of choosing which single one of the $6$ factors donates a $y$. Any one of the six could be the chosen factor.

$\binom{6}{1} = 6$

This is why the second entry of every row of Pascal's triangle equals the row number, and it is a fast way to check you are reading the right row.

Every term comes from choosing a $y$ in $k$ of the $n$ factors and an $x$ in the other $n - k$, giving $x^{n-k}y^k$, and there are $\binom{n}{k}$ ways to make that choice.

$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k}\, x^{n-k} y^{k}$

The sum must start at $k = 0$ to include the $x^n$ term, the exponents must add to $n$ (which rules out $x^ky^k$), and the coefficient $\binom{n}{k}$ cannot be dropped, since $(x+y)^2$ is $x^2 + 2xy + y^2$, not $x^2 + xy + y^2$.

Use the theorem with $A = x$, $B = 2$ and row $3$, which is $1,\ 3,\ 3,\ 1$:

$(x + 2)^3 = x^3 + 3x^2(2) + 3x(2)^2 + (2)^3$

Evaluate the powers of $2$ before multiplying: $2^2 = 4$ and $2^3 = 8$.

$(x + 2)^3 = x^3 + 6x^2 + 12x + 8$

Check with $x = 1$: the left side is $3^3 = 27$, and the right side is $1 + 6 + 12 + 8 = 27$.

The power $(x + y)^n$ is a product of $n$ identical factors, and each factor donates exactly one letter, either an $x$ or a $y$.

So a term is built from $n$ letters in total. If $k$ of them are y's, the other $n - k$ are x's:

$(n - k) + k = n$

Every term of $(x+y)^7$ therefore has exponents summing to $7$, which is why there is no $x^4y^2$ term to look for.