4.1 Binomial expansion for positive integer n; n! and nCr — Practise with help

1. Add the indices for a product: $x^3 \times x^4 = x^7$. Subtract them for a quotient: $\dfrac{x^7}{x^3} = x^4$.
2. $(x + 3)(x - 2) = x^2 - 2x + 3x - 6 = x^2 + x - 6$.
3. $3^4 = 81$ and $(-2)^3 = -8$.

If question 3 was hard, revisit the laws of indices in Section 2.1 before you continue. An odd power of a negative number is negative.

The factorial multiplies every whole number from $1$ up to that number. For the coefficient, use $\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$ with $n = 6$ and $r = 2$.

Work out $6!$ and the two factorials on the bottom, $2!$ and $4!$. The $4!$ appears on the top and the bottom, so it cancels.

There are $15$ ways to choose $2$ items from $6$.

Use row 4 of Pascal's triangle. Its entries are the coefficients $1, 4, 6, 4, 1$.

Write the powers of $x$ in order from $x^0$ to $x^4$, then put each coefficient in front of its power.

The constant term is $1^4 = 1$, and the coefficient of $x^4$ is $1$, which matches the outer entries of the row.

Start from the answer to question 2 and replace each $x$ with $3x$. Take care with the powers of $3$.

Work out $(3x)^2 = 9x^2$, $(3x)^3 = 27x^3$ and $(3x)^4 = 81x^4$ before you write the terms out.

The coefficients grow because each power of $3x$ brings in a power of $3$.

Use the single-term form $\binom{n}{r}a^{n-r}(bx)^r$, with $n = 5$, $r = 2$, $a = 3$ and $b = 2$.

Work out the three numbers separately: $\binom{5}{2} = 10$, then $3^3 = 27$, then $2^2 = 4$.

The term is $1080x^2$, so the coefficient is $1080$. The factor $3^3 = 27$ comes from the part of the bracket that does not contain $x$.

Here $b = -x$, so every odd power of $x$ carries a minus sign. Use the coefficients $1, 5, 10, 10, 5, 1$.

Attach the powers of $-x$ to the powers of $2$ one term at a time. For the third term, $\binom{5}{2}2^3(-x)^2 = 10 \times 8 \times x^2$.

The constant term is $2^5 = 32$, and the final term is $(-x)^5 = -x^5$. The signs alternate because each term brings one more factor of $-x$.

Here $a = 1$, so the power of $a$ is always $1$ and does not change the coefficient. You need the term with $r = 3$.

Work out $\binom{7}{3} = 35$, then $(-2)^3 = -8$. The coefficient is the product of those two numbers.

The term is $-280x^3$. It is negative because the power of $-2x$ is odd.

Write the value as $800(1 + 0.05)^2$, then expand the bracket with $n = 2$ and the coefficients $1, 2, 1$.

Expand $(1 + 0.05)^2 = 1 + 2(0.05) + (0.05)^2 = 1 + 0.1 + 0.0025$.

so

The account holds £882 after 2 years. The extra £82 is the two years of interest added to the original £800.