IB Maths AA HL 1.3 Geometric sequences and series

[Maximum number: 5]

In this question you will investigate series of the form

\sum_{i=1}^{n} \boldsymbol{i}^{q}=1^{q}+2^{q}+3^{q}+\ldots+n^{q} \text { where } n, q \in \mathbb{Z}^{+}

and use various methods to find polynomials, in terms of n, for such series.
When q=1, the above series is arithmetic.

(a)

By considering $f(x)=1+x+x^{2}+\ldots+x^{n}$ as a geometric series, for $x \neq 1$ , show that $f(x)=\frac{x^{n+1}-1}{x-1}$ .

[ 2 ]

uses $S_{n}=\frac{u_{1}\left(r^{n}-1\right)}{r-1}$ with r=x and $u_{1}=1$
clear indication there are ( n+1 ) terms

f(x)=\frac{x^{n+1}-1}{x-1}

(b)

For $x \neq 1$ , show that $f_{1}(x)=\frac{n x^{n+2}-(n+1) x^{n+1}+x}{(x-1)^{2}}$ .

[ 3 ]

METHOD 1

\begin{aligned} & f_{1}(x)=x f^{\prime}(x) \\ & =x \frac{(x-1)(n+1) x^{n}-1 \times\left(x^{n+1}-1\right)}{(x-1)^{2}} \end{aligned}

Note: Award M1 for attempting to use the quotient or the product rule to find $f^{\prime}(x)$ .

=x \frac{(n x+x-n-1) x^{n}-\left(x^{n+1}-1\right)}{(x-1)^{2}}\left(=x \frac{n x^{n+1}-n x^{n}-x^{n}+1}{(x-1)^{2}}\right)

Note: Award A1 for any correct manipulation of the derivative that leads to the AG.

=\frac{n x^{n+2}-(n+1) x^{n+1}+x}{(x-1)^{2}}

METHOD 2
attempts to form $(x-1) f_{1}(x)$

\begin{aligned} & (x-1) f_{1}(x)=n x^{n+1}-\left(x+x^{2}+x^{3}+\ldots+x^{n}\right) \\ & f_{1}(x)=\frac{1}{x-1}\left(n x^{n+1}-\left(\frac{x^{n+1}-1}{x-1}-1\right)\right) \\ & f_{1}(x)=\frac{1}{x-1}\left(\frac{n x^{n+1}(x-1)-x^{n+1}+x}{x-1}\right)\left(=\frac{1}{x-1}\left(\frac{n x^{n+2}-n x^{n+1}-x^{n+1}+x}{x-1}\right)\right) \end{aligned}

Note: Award A1 for any correct manipulation of the derivative that leads to the AG.

f_{1}(x)=\frac{n x^{n+2}-(n+1) x^{n+1}+x}{(x-1)^{2}}

IB Maths AA HL Sl 1 3 Geometric Sequences and Series Questions