Edexcel AEA 2016 June — Question 5 13 marks

Exam BoardEdexcel
ModuleAEA (Advanced Extension Award)
Year2016
SessionJune
Marks13
PaperDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeGeometric series with summation
DifficultyChallenging +1.8 This AEA question requires differentiating a geometric series to find a weighted sum, then applying the result with substitution. While it involves multiple sophisticated steps (algebraic manipulation of series, differentiation technique, substitution and simplification), the path is relatively guided through parts (a)-(c). The techniques are advanced but systematic rather than requiring deep insight, placing it above typical A-level but below the most challenging AEA problems.
Spec1.04g Sigma notation: for sums of series1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1

5.(a)Show that $$\sum _ { r = 0 } ^ { n } x ^ { - r } = \frac { x } { x - 1 } - \frac { x ^ { - n } } { x - 1 } \quad \text { where } x \neq 0 \text { and } x \neq 1$$ (b)Hence find an expression in terms of \(x\) and \(n\) for \(\sum _ { r = 0 } ^ { n } r x ^ { - ( r + 1 ) }\) for \(x \neq 0\) and \(x \neq 1\) Simplify your answer.
(c)Find \(\sum _ { r = 0 } ^ { n } \left( \frac { 3 + 5 r } { 2 ^ { r } } \right)\) Give your answer in the form \(a - \frac { b + c n } { 2 ^ { n } }\) ,where \(a , b\) and \(c\) are integers.

5.(a)Show that

$$\sum _ { r = 0 } ^ { n } x ^ { - r } = \frac { x } { x - 1 } - \frac { x ^ { - n } } { x - 1 } \quad \text { where } x \neq 0 \text { and } x \neq 1$$

(b)Hence find an expression in terms of $x$ and $n$ for $\sum _ { r = 0 } ^ { n } r x ^ { - ( r + 1 ) }$ for $x \neq 0$ and $x \neq 1$\\
Simplify your answer.\\
(c)Find $\sum _ { r = 0 } ^ { n } \left( \frac { 3 + 5 r } { 2 ^ { r } } \right)$

Give your answer in the form $a - \frac { b + c n } { 2 ^ { n } }$ ,where $a , b$ and $c$ are integers.\\

\hfill \mbox{\textit{Edexcel AEA 2016 Q5 [13]}}