FP1 June 2016 Q8
8.
| Scheme | Marks |
|---|---|
| If \(n = 1\), \(\displaystyle\sum_{r=1}^{n} \frac{2r + 1}{r^2(r + 1)^2} = \frac{3}{4}\) and \(1 - \dfrac{1}{(n + 1)^2} = \dfrac{3}{4}\), so true for \(n = 1\). | B1 |
| Assume result true for \(n = k\) and consider \(\displaystyle\sum_{r=1}^{k+1} \frac{2r + 1}{r^2(r + 1)^2} = 1 - \frac{1}{(k + 1)^2} + \frac{2(k + 1) + 1}{(k + 1)^2(k + 2)^2}\) | M1 |
| \(= 1 - \left(\dfrac{(k + 2)^2}{(k + 1)^2(k + 2)^2} - \dfrac{2(k + 1) + 1}{(k + 1)^2(k + 2)^2}\right) = 1 - \left(\dfrac{(k^2 + 2k + 1)}{(k + 1)^2(k + 2)^2}\right)\) | A1 |
| \(= 1 - \left(\dfrac{(k + 1)^2}{(k + 1)^2(k + 2)^2}\right) = 1 - \left(\dfrac{1}{(k + 1 + 1)^2}\right)\) | M1 |
| True for \(\boldsymbol{n = k + 1}\) if true for \(\boldsymbol{n = k}\), (and true for \(\boldsymbol{n = 1}\)) so true by induction for all \(\boldsymbol{n \in \mathbb{Z}^{+}}\) | A1cso |
| (5) |
Notes
B1: Checks \(n = 1\) on both sides and states true for \(n = 1\) seen anywhere
M1: (Assumes true for) \(n = k\) and adds \((k+1)^{\text{th}}\) term to sum of \(k\) terms
A1: \(1 - \left(\dfrac{(k^2 + 2k + 1)}{(k + 1)^2(k + 2)^2}\right)\) seen (linked to 2nd M)
M1: \((k + 1)^2(k + 2)^2\) attempted as common denominator of two fractions.
A1cso: Makes correct complete induction statement including at least statements in bold. Accept \(n \geqslant 1\) or \(n = 1, 2, 3\ldots\) or all positive Integers or all \(n\). Statement true for \(n = 1\) here could contribute to B1 mark earlier.
| Scheme | Marks |
|---|---|
| \(n = 1\): \(u_1 = 5 \times \left(\tfrac{1}{3}\right)^1 + \tfrac{4}{3} = 3\) so expression for \(u_n\) true for \(n = 1\) | B1 |
| Assume result true for \(n = k\) and consider \(u_{k+1} = \tfrac{1}{3}\left(5 \times \left(\tfrac{1}{3}\right)^k + \tfrac{4}{3}\right) + \tfrac{8}{9}\) | M1 |
| Obtain \(u_{k+1} = 5 \times \left(\tfrac{1}{3}\right)^{k+1} + \tfrac{4}{9} + \tfrac{8}{9}\) | A1 |
| \(5 \times \left(\dfrac{1}{3}\right)^{k+1} + \dfrac{4}{3}\) and deduce that result is true for \(n = k + 1\) | dM1 |
| True for \(\boldsymbol{n = k + 1}\) if true for \(\boldsymbol{n = k}\), (and true for \(\boldsymbol{n = 1}\)) so true by induction for all \(\boldsymbol{n \in \mathbb{Z}^{+}}\) | A1 cso |
| (5) | |
| (10 marks) |
Notes
B1: Checks \(n = 1\) in \(u_n\) and states true for \(n = 1\) seen anywhere.
M1: (Assumes result for) \(n = k\) and substitutes \(u_k\) into correct expression for \(u_{k+1}\)
A1: \(\dfrac{4}{9} + \dfrac{8}{9}\) or \(\dfrac{1}{3} \cdot \dfrac{4}{3} + \dfrac{8}{9}\) seen
dM1: Obtains \(5 \times \left(\dfrac{1}{3}\right)^{k+1} + \dfrac{4}{3}\) and statement true for \(n = k + 1\) or equivalent seen anywhere dependent on previous M.
A1cso: Makes correct complete induction statement including at least statements in bold. Accept \(n \geqslant 1\) or \(n = 1, 2, 3\ldots\) or all positive Integers or all \(n\). Statement true for \(n = 1\) here could contribute to B1 mark earlier.