A series is given by $\sum_{r=1}^{n} 2(4-3x)^r$ where $x$ is constant. Explain why this is

Question

A series is given by $\sum_{r=1}^{n} 2(4-3x)^r$ where $x$ is constant. Explain why this is a geometric series. Determine the range of values of $x$ for the sum to infinity of this series to exist.[3] Using $x=1$, and given that $\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)$, find $\sum_{r=0}^{n-1} [2(4-3x)^r + (r+1)(2r+5)]$, leaving your answer in the form of $n(an^2+bn+c)$, where $a$, $b$ and $c$ are constants to be determined.[4]

Accepted Answer

(a) $1<x<\dfrac53$ (b) $n\left(\dfrac23n^2+\dfrac52n+\dfrac{23}{6}\right)$, so $a=\dfrac23,\ b=\dfrac52,\ c=\dfrac{23}{6}$

2025 DHS P1 Q2

2025 DHS P1 Q2

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