# Question 8:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| If $n=1$: $\sum_{r=1}^{1}\frac{2r+1}{r^2(r+1)^2} = \frac{3}{4}$ and $1-\frac{1}{(1+1)^2}=\frac{3}{4}$, so true for $n=1$ | B1 | Checks $n=1$ on both sides and states true for $n=1$ |
| Assume true for $n=k$ and consider $\sum_{r=1}^{k+1}\frac{2r+1}{r^2(r+1)^2} = 1-\frac{1}{(k+1)^2}+\frac{2(k+1)+1}{(k+1)^2(k+2)^2}$ | M1 | Assumes true for $n=k$ and adds $(k+1)$th term |
| $= 1-\left(\frac{(k+2)^2}{(k+1)^2(k+2)^2}-\frac{2(k+1)+1}{(k+1)^2(k+2)^2}\right) = 1-\left(\frac{k^2+2k+1}{(k+1)^2(k+2)^2}\right)$ | A1 | $(k+1)^2(k+2)^2$ as common denominator attempted |
| $= 1-\left(\frac{(k+1)^2}{(k+1)^2(k+2)^2}\right) = 1-\left(\frac{1}{(k+1+1)^2}\right)$ | M1 | |
| **True** for $n=k+1$ if **true** for $n=k$, (and **true** for $n=1$) so **true** by induction for all $n \in \mathbb{Z}^+$ | A1 cso | Complete induction statement with bold statements; accept $n\geq1$ or $n=1,2,3\ldots$ |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1$: $u_1 = 5\times(\frac{1}{3})^1+\frac{4}{3}=3$, so true for $n=1$ | B1 | Checks $n=1$ in $u_n$ and states true |
| Assume true for $n=k$ and consider $u_{k+1} = \frac{1}{3}(5\times(\frac{1}{3})^k+\frac{4}{3})+\frac{8}{9}$ | M1 | Assumes result for $n=k$ and substitutes $u_k$ into correct $u_{k+1}$ expression |
| Obtain $u_{k+1} = 5\times(\frac{1}{3})^{k+1}+\frac{4}{9}+\frac{8}{9}$ | A1 | $\frac{4}{9}+\frac{8}{9}$ or $\frac{1}{3}\cdot\frac{4}{3}+\frac{8}{9}$ seen |
| $5\times\left(\frac{1}{3}\right)^{k+1}+\frac{4}{3}$ and deduce true for $n=k+1$ | dM1 | Obtains $5\times(\frac{1}{3})^{k+1}+\frac{4}{3}$; dependent on previous M |
| **True** for $n=k+1$ if **true** for $n=k$, (and **true** for $n=1$) so **true** by induction for all $n \in \mathbb{Z}^+$ | A1 cso | Complete induction statement; accept $n\geq1$ or $n=1,2,3\ldots$ |

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