Practice section 9 of 17

In an infinite geometric progression with a nonzero first term and common ratio $\frac1{\sqrt[6]{2}}$, the sum of the first $n$ terms equals the sum of all the remaining terms. Find $n$.

Given $\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}6$ and $\sum_{r=1}^{n}r^3=[\frac n2(n+1)]^2$, evaluate: (a) $1(1)+2(3)+3(5)+\cdots+n(2n-1)$; (b) $\sum_{r=0}^{n}(ar^3+a^r-a)$ in terms of $a$ (and $n$); (c) $(n+1)^2+(n+2)^2+\cdots+(2n)^2$.

A poultry farmer raises chickens, ducks and geese. Allowing for conditions such as sales and deaths from infection, the geese population at the end of $n$ months is modelled by $u_n$, with $u_{n+1}=0.75u_n+80$ and initial population $u_0$. Calculate $u_1,u_2,u_3,u_4$ for (i) $u_0=576$ and (ii) $u_0=320$, commenting briefly on the results. Find $u_n$ in terms of $n,u_0$; deduce that $u_n\to320$ for large $n$, and use the formula to justify the earlier comments.

A sequence satisfies $u_{n+1}=2u_n+An$ for $n\ge1$, where $A$ is constant.

(i) Given $u_1=5,u_2=15$, find $A$ and $u_3$.

Its nth term has the form $u_n=a2^n+bn+c$, where $a,b,c$ are constants.

(ii) Find $a,b,c$.

(iii) Find $\sum_{r=1}^{n}u_r$ in terms of $n$; you need not simplify your answer.

The graph is $y=e^x-3x$, whose roots are $\alpha<\beta$.

(i) Find both roots to 3 decimal places.

A real sequence satisfies $x_{n+1}=\frac13e^{x_n}$ for $n\ge1$.

(ii) Prove algebraically that, if it converges, its limit is $\alpha$ or $\beta$.

(iii) Use a calculator to determine its behaviour for $x_1=0,1,2$.

(iv) By considering $x_{n+1}-x_n$, prove that $x_{n+1}<x_n$ if $\alpha<x_n<\beta$, and $x_{n+1}>x_n$ if $x_n<\alpha$ or $x_n>\beta$.

(v) Relate (iv) to the behaviours in (iii).

A radius-$2$ circle is cut into $10$ unequal sectors with areas in arithmetic progression. The first and smallest has area $a$, followed by $a+d,a+2d,\ldots$ square units. Show that $2a+9d=\frac{4\pi}{5}$. If the first, fourth and ninth sector areas are consecutive GP terms, show that $a=\frac\pi5$. Hence find the area of the largest sector.