FP1 2016 8 Edexcel

8

(i) At n = 1 we have:

\displaystyle \sum_{r = 1}^1 \frac {2r + 1} {r^2 (r + 1)^2} = \frac {2 + 1} {2^2} = \frac 3 4 = 1 - \frac 1 4 = 1 - \frac 1 {(1 + 1)^2}

So the formula holds for n = 1. Suppose it holds for some n. Then:

\begin{align*}\sum_{r = 1}^{n + 1} \frac {2r + 1} {r^2(r + 1)^2} & = \frac {2(n + 1) + 1} {(n + 1)^2(n + 2)^2} + \sum_{r = 1}^n \frac {2r + 1} {r^2 (r + 1)^2} \\ & = \frac {2n + 3} {(n + 1)^2(n + 2)^2} + 1 - \frac 1 {(n + 1)^2} \\ & = 1 + \frac {(2n + 3) - (n + 2)^2} {(n + 1)^2 (n + 2)^2} \\ & = 1 + \frac {(2n + 3) - (n^2 + 4n + 4)} {(n + 1)^2 (n + 2)^2} \\ & = 1 - \frac {n^2 + 2n + 1} {(n + 1)^2 (n + 2)^2} \\ & = 1 - \frac {(n + 1)^2} {(n + 1)^2 (n + 2)^2} \\ & = 1 - \frac 1 {(n + 1 + 1)^2}\end{align*}

So if the formula holds for n it holds for n + 1. Since it holds for n = 1 the sum holds for all positive integers n by induction.

(ii)

We have:

\displaystyle 5 \left(\frac 1 3\right)^1 + \frac 4 3 = \frac 5 3 + \frac 4 3 = 3 = u_1

So the formula works for n = 1. Suppose it works for some n. Then:

\begin{align*}\frac 1 3 u_n + \frac 8 9 & = \frac 1 3 \left(5 \times \left(\frac 1 3\right)^n + \frac 4 3\right) + \frac 8 9 \\ & = 5 \times \left(\frac 1 3\right)^{n + 1} + \frac 4 9 + \frac 8 9 \\ & = 5 \times \left(\frac 1 3\right)^{n + 1} + \frac {12} 9 \\ & = 5 \times \left(\frac 1 3\right)^{n + 1} + \frac 4 3 \\ & = u_{n + 1} \end{align*}

So if the formula holds for some n, it holds for n + 1. Since it holds for n = 1, it holds for all positive n by induction.

Alternative to part (i): We can notice that:

\displaystyle \frac {2r + 1} {r^2 (r + 1)^2} = \frac 1 {r^2} - \frac 1 {(r + 1)^2}

while we could notice that the series telescopes, this wouldn’t use induction. But in the induction step it simplifies the calculation a bit, since we’d simply have:

\displaystyle \frac 1 {(n + 1)^2} - \frac 1 {(n + 2)^2} + 1 - \frac 1 {(n + 1)^2} = 1 - \frac 1 {(n + 1)^2}

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