Expand $\sum\limits_{r = 4}^7 {{2^r}}$ as the sum of four terms.

(i)     Find the value of $\sum\limits_{r = 4}^{30} {{2^r}}$ .

(ii)    Explain why $\sum\limits_{r = 4}^\infty  {{2^r}}$ cannot be evaluated.

$\sum\limits_{r = 4}^7 {{2^r}}  = {2^4} + {2^5} + {2^6} + {2^7}$ (accept $16 + 32 + 64 + 128$ )     A1     N1

[1 mark]

(i) METHOD 1

recognizing a GP     (M1)

${u_1} = {2^4}$ , $r = 2$ , $n = 27$     (A1)

correct substitution into formula for sum     (A1)

e.g. ${S_{27}} = \frac{{{2^4}({2^{27}} - 1)}}{{2 - 1}}$

${S_{27}} = 2147483632$     A1     N4

METHOD 2

recognizing $\sum\limits_{r = 4}^{30} { = \sum\limits_{r = 1}^{30} { - \sum\limits_{r = 1}^3 {} } }$     (M1)

recognizing GP with ${u_1} = 2$ , $r = 2$ , $n = 30$     (A1)

correct substitution into formula for sum

${S_{30}} = \frac{{2({2^{30}} - 1)}}{{2 - 1}}$     (A1)

$= 214783646$

$\sum\limits_{r = 4}^{30} {{2^r}}  = 2147483646 - (2 + 4 + 8)$

$= 2147483632$     A1     N4

(ii) valid reason (e.g. infinite GP, diverging series), and $r \ge 1$ (accept $r > 1$ )     R1R1     N2

[6 marks]

This question proved difficult for many candidates. A number of students seemed unfamiliar with sigma notation. Many were successful with part (a), although some listed terms or found an overall sum with no working.

The results for part (b) were much more varied. Many candidates did not realize that $n$ was $27$ and used $30$ instead. Very few candidates gave a complete explanation why the infinite series could not be evaluated; candidates often claimed that the value could not be found because there were an infinite number of terms.