Sequences & Series (Edexcel International A Level (IAL) Maths: Pure 2): Exam Questions

The first  $k$ terms of a series are given by $\sum _{r=1}^k\left(7+5r\right)$

Show that this is an arithmetic series, and determine its first term and common difference.

 Given that $\sum _{r=1}^k\left(7+5r\right)$ $=1190$, 

(i) Show that $(5k+119)(k-20)=0$

(ii) Hence find the value of $k$.

The first $k$ terms of a series are given by $\sum _{r=1}^{k}5\times 2^r$. 

Show that this is a geometric series, and determine its first term and common ratio.

Given that $\sum _{r=1}^{k}5\times 2^r=20470$, 

Show that $k=\frac{\log 2048}{\log 2}$

For this value of $k$, calculate $\sum _{r=1}^{k+3}5\times 2^r$.

A geometric series is given by $1+2x+4x^2+...$

Write down the common ratio, $r$ , of the series.

Given that the series is convergent, and that  $\sum _{n=1}^{\infty }{\left(2x\right)}^{n-1}$$=19$, calculate the value of $x$.

An arithmetic series is given by $a+(a+d)+(a+2d)+...$

Given that  $\sum _{n=1}^7\left(a+\left(n-1\right)d\right)=91$ and  $\sum _{n=1}^{10}\left(a+\left(n-1\right)d\right)=175$, find the values of $a$ and $d$.

The terms of a sequence are defined by $u_k=k^2$  for all $k\ge 1$. 

State, with a reason, whether this sequence is increasing, decreasing, or neither.

 It can be shown that, for all $n\ge 1$,

               $\sum _{r=1}^n{r}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ 

Using that formula,

Calculate $\sum _{r=1}^{50}{u}_r$        

Find the value of  ${51}^2+{52}^2+{53}^2+...$$+{99}^2+{100}^2$, i.e. the sum of the squares of all the integers between 51 and 100 inclusive.

Given that $\sum _{r=1}^k\left(31-6r\right)=-943$, 

(i)

Show that $(3k+41)(k-23)=0$

(ii)

Hence, find the value of $k$.   

 Given that  $\sum _{n=1}^9\left(a+\left(n-1\right)d\right)=-279$ and  $\sum _{n=1}^{13}\left(a+\left(n-1\right)d\right)=-585$, find the values of  $a$ and $d$ .

Given that  $\sum _{r=1}^{k}7\times 3^r=620004$, 

Show that  $k=\frac{\log 59049}{\log 3}$

For this value of $k$, calculate $\sum _{r=0}^{k+3}7\times 3^r$.

A convergent geometric series is given by $1-4x+16x^2-64x^3+...$ 

Write down the range of possible values of $x$.

Given that $\sum _{n=1}^{\infty }{\left(-4x\right)}^{n-1}=24$    

Calculate the value of $x$.

The terms of a sequence are defined, for all $k\ge 1,$ by  $u_k=(-1{)}^k\times k^2$ . 

State, with a reason, whether this sequence is increasing, decreasing, or neither.

It can be shown that, for all $n\ge 1$, 

       $\sum _{r=1}^n{\left(2r\right)}^2=\frac{2n\left(n+1\right)\left(2n+1\right)}{3}$ and $\sum _{r=1}^n{\left(2r-1\right)}^2=\frac{n\left(2n+1\right)\left(2n-1\right)}{3}$

Using those formulas,

Show that $\sum _{r=1}^{100}{u}_r=\sum _{r=1}^{100}r$.

Given that  $\sum _{r=1}^k\left(89-5r\right)=-35$, find the value of $k$.

Given that  $\sum _{r=1}^{k}3\times {\left(-2\right)}^r=-262146$, 

(i)   show that  $\frac{k-1}{2}=\frac{\log 65536}{\log 4}$    

(ii)   hence find the value of $k$.

For this value of $k$, calculate $\sum _{r=5}^{k+2}3\times {\left(-2\right)}^r$.

Given that  $\sum _{n=7}^{12}\left(a+\left(n-1\right)d\right)=-69$,  $\sum _{n=7}^{16}\left(a+\left(n-1\right)d\right)=-175$, and $\sum _{n=1}^6\left(a+\left(n-1\right)d\right)=-13d$, find the values of $a$ and $d$.

A convergent geometric series is given by $\sqrt{3}+\sqrt{6x}+2x\sqrt{3}+...$, where in all cases the square root symbol indicates the positive square root of the number in question. 

Write down the range of possible values of $x$.

Given that $\sum _{n=2}^{\infty }\sqrt{3}\times {\left(\sqrt{2x}\right)}^{n-1}=3\sqrt{3}$      

Calculate the value of $x$.

A sequence is defined for $k\ge 1$ by $u_k=\sqrt{13}+{\left(-2\right)}^{k-1}$ .

Calculate $\sum _{r=11}^{23}{u}_r$ , giving your answer as an exact value.

A sequence is defined for all $k\ge 1$ by

                 $u_k=-2k\times {\left(\cos \left(k\pi \right)\right)}^{k+1}$ 

Determine, giving reasons for your answer, whether the sequence is increasing, decreasing, or neither.

A different sequence is defined for all $k\ge 1$ by

               $v_k=\sin \left(kq\mathrm{\pi }\right)$

where $q$ is a real constant. 

Given that the sequence is not periodic, 

suggest a possible value for $q$, giving a reason for your answer.

Prove that, for all $n\ge 1$,

         $\sum _{r=1}^n{\left(2r\right)}^2-\sum _{r=1}^n{\left(2r-1\right)}^2=\sum _{r=1}^{2n}r$