SQA Advanced Higher Mathematics Applications of Algebra and Calculus overview quiz quiz

$\dfrac{n!}{r!}$

$\dfrac{n!}{r!(n - r)!}$

$\dfrac{(n - r)!}{r!}$

$\dfrac{r!}{n!}$

$r$

$n - r$

$n$

$n + 1$

$1$

$n$

$0$

$r$

$ar^{n - 1}$

$a + (n - 1)d$

$\dfrac{n}{2}(2a + (n - 1)d)$

$a + nd$

$r > 1$

$|r| < 1$

$r = 1$

$r$ is an integer

$14$

$16$

$32$

It diverges

$x = 1$

$x = 0$

$x = a$

the stationary point

$f^{(n)}(0)$

$\dfrac{f^{(n)}(0)}{n!}$

$n! \, f^{(n)}(0)$

$\dfrac{f(0)}{n}$

$1 - x + \dfrac{x^2}{2} - \cdots$

$x - \dfrac{x^3}{6} + \cdots$

$1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \cdots$

$1 + \dfrac{x^2}{2} + \dfrac{x^4}{24} + \cdots$

$\dfrac{n(n + 1)}{2}$

$\dfrac{n(n + 1)(2n + 1)}{6}$

$n^2$

$\dfrac{n^2(n + 1)^2}{4}$

$\dfrac{n(n + 1)}{2}$

$\dfrac{n(n + 1)(2n + 1)}{6}$

$\dfrac{n^2(n + 1)^2}{4}$

$n^3$

The inductive step

The assumption

The base case or concluding statement

The algebra

The numerator is zero

The denominator is zero and the numerator is not

The function has a maximum

The degree of the numerator exceeds the denominator

The numerator and denominator have equal degree

The numerator degree is one more than the denominator

The denominator has no real roots

The function is even

$f(-x) = f(x)$

$f(-x) = -f(x)$

$f(0) = 0$

it has a stationary point at the origin