Summation of finite seriesEdexcel International A Level Further Maths: Subtopic test

1. 1
   A sequence has $r$th term $u_r=2r+3$, and $S_n$ is the sum of the first $n$ terms.

   (a)

   Find an expression for $S_n$.

   [1 mark]

   • A$n^2+n+3$
   • B$\frac{n(n+1)}{2}+3n$
   • C$n(n+4)$
   • D$2n(n+1)+3n$

   (b)

   Find the value of $S_{20}$.

   [1 mark]

   • A$480$
   • B$423$
   • C$270$
   • D$440$

   (c)

   Find the smallest value of $n$ for which $S_n>1000$.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Let $S_n=1\times3+2\times4+3\times5+\dots+n(n+2)$, the sum of the first $n$ terms of a series.

   (a)

   Find a simplified expression for $S_n$.

   [1 mark]

   • A$\frac{n(n+1)(n+2)}{3}$
   • B$\frac{n(n+1)(2n+7)}{6}$
   • C$\frac{n(n+1)(2n+1)}{6}+2$
   • D$\frac{n^2(n+1)^2(2n+1)}{6}$

   (b)

   Find the value of $S_{10}$.

   [1 mark]

   • A$385$
   • B$440$
   • C$110$
   • D$495$

   (c)

   Find the sum of the 6th to the 10th terms of the series, inclusive.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   An orange display is built in layers. Layer $r$, counting from the top with $r=1,2,3,\dots$, is a square arrangement containing $r^2$ oranges.

   (a)

   Find the number of oranges in layers $13$ to $20$ inclusive.

   [3 marks]

   (b)

   A second display has $r(r+3)$ oranges in layer $r$. Show that the total number of oranges in $n$ layers of the second display is $\frac{n(n+1)(n+5)}{3}$.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A sequence has $r$th term $u_r=r(2r+3)$, and $S_n=\sum_{r=1}^{n}u_r$.

   (a)

   (i) Show that $S_n=\frac{n(n+1)(4n+11)}{6}$.
   (ii) Hence find the sum of the 11th to the 20th terms of the sequence.

   [6 marks]

   (b)

   Given that $S_n=1210$, show that $4n^3+15n^2+11n-7260=0$ and hence find the value of $n$, justifying that there is only one possible value.

   [6 marks]

   Total for question 4: 12 marks

End of questions