Standard +0.3 This is a standard proof by induction for a summation formula with a straightforward algebraic structure. The partial fractions decomposition (1/((2n-1)(2n+1)) = 1/2(1/(2n-1) - 1/(2n+1))) makes the telescoping pattern clear, and the inductive step involves routine algebraic manipulation. While it requires understanding of the induction framework and careful algebra, it's a textbook example without novel insight, making it slightly easier than average for A-level.
Therefore if true for \(n=k\) it is also true for \(n=k+1\)
E1
Dependent on A1 and previous E1
Since it is true for \(n=1\), it is true for all positive integers \(n\)
E1 [7]
Dependent on B1 and previous E1. E0 if "last term" = "sum of terms" seen above
## Question 6:
When $n=1$: $\dfrac{1}{1\times 3} = \dfrac{1}{3}$ and $\dfrac{n}{2n+1} = \dfrac{1}{3}$, so true for $n=1$ | B1 | Condone e.g. "$\dfrac{1}{3}=\dfrac{1}{3}$"
Assume true for $n=k$ | E1 | Assuming true for $k$; statement of assumed result not essential but further work should be seen
Sum of $k+1$ terms $= \dfrac{k}{2k+1} + \dfrac{1}{(2k+1)(2k+3)}$ | M1 | Adding correct $(k+1)$th term to sum for $k$ terms
$= \dfrac{k(2k+3)+1}{(2k+1)(2k+3)}$ | M1 | Combining fractions
$= \dfrac{2k^2+3k+1}{(2k+1)(2k+3)} = \dfrac{(k+1)(2k+1)}{(2k+1)(2k+3)} = \dfrac{k+1}{2k+3}$ | A1 | Complete accurate work
which is $\dfrac{n}{2n+1}$ with $n=k+1$
Therefore if true for $n=k$ it is also true for $n=k+1$ | E1 | Dependent on A1 and previous E1
Since it is true for $n=1$, it is true for all positive integers $n$ | E1 **[7]** | Dependent on B1 and previous E1. E0 if "last term" = "sum of terms" seen above
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