Standard +0.8 This is a standard proof by induction for a summation involving exponentials and a linear term (r·3^(r-1)). While it requires careful algebraic manipulation in the inductive step—particularly factoring out 3^n and simplifying to match the target form—it follows the standard induction template without requiring novel insight. The algebraic complexity is moderate but manageable for Further Maths students, placing it somewhat above average difficulty.
When \(n=1\): \(\sum_{r=1}^{1} r \cdot 3^{r-1} = 1 \times 3^0 = 1\) and \(\tfrac{1}{4}[3^n(2n-1)+1] = \tfrac{1}{4}[3(2-1)+1] = 1\), so true for \(n=1\)
B1
Assume \(\sum_{r=1}^{k} r \cdot 3^{r-1} = \tfrac{1}{4}[3^k(2k-1)+1]\)
E1
Assuming true for \(k\)
\(\sum_{r=1}^{k+1} r \cdot 3^{r-1} = \tfrac{1}{4}[3^k(2k-1)+1] + (k+1)3^{k+1-1}\)
M1*
Adding \((k+1)\)th term (incorrect expressions on LHS lose final E1)
\(= \tfrac{1}{4}[3^k(2k-1)+1+4(k+1)3^k]\)
M1 dep*
Attempt to obtain factor of \(\tfrac{1}{4}\)
\(= \tfrac{1}{4}[3^k(2k-1+4(k+1))+1]\)
M1 dep*
For \([3^k(ak+b)+c]\), \(c \neq 0\)
\(= \tfrac{1}{4}[3^k(6k+3)+1]\)
\(= \tfrac{1}{4}[3^{k+1}(2k+1)+1]\)
A1
\(= \tfrac{1}{4}[3^{k+1}(2(k+1)-1)+1]\)
Or target seen
Therefore if true for \(n=k\) it is also true for \(n=k+1\). Since it is true for \(k=1\), it is true for all positive integers.
E1, E1
Dependent on A1 and previous E1; Dependent on B1 and previous E1
[8]
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $n=1$: $\sum_{r=1}^{1} r \cdot 3^{r-1} = 1 \times 3^0 = 1$ and $\tfrac{1}{4}[3^n(2n-1)+1] = \tfrac{1}{4}[3(2-1)+1] = 1$, so true for $n=1$ | B1 | |
| Assume $\sum_{r=1}^{k} r \cdot 3^{r-1} = \tfrac{1}{4}[3^k(2k-1)+1]$ | E1 | Assuming true for $k$ |
| $\sum_{r=1}^{k+1} r \cdot 3^{r-1} = \tfrac{1}{4}[3^k(2k-1)+1] + (k+1)3^{k+1-1}$ | M1* | Adding $(k+1)$th term (incorrect expressions on LHS lose final E1) |
| $= \tfrac{1}{4}[3^k(2k-1)+1+4(k+1)3^k]$ | M1 dep* | Attempt to obtain factor of $\tfrac{1}{4}$ |
| $= \tfrac{1}{4}[3^k(2k-1+4(k+1))+1]$ | M1 dep* | For $[3^k(ak+b)+c]$, $c \neq 0$ |
| $= \tfrac{1}{4}[3^k(6k+3)+1]$ | | |
| $= \tfrac{1}{4}[3^{k+1}(2k+1)+1]$ | A1 | |
| $= \tfrac{1}{4}[3^{k+1}(2(k+1)-1)+1]$ | | Or target seen |
| Therefore if true for $n=k$ it is also true for $n=k+1$. Since it is true for $k=1$, it is true for all positive integers. | E1, E1 | Dependent on A1 and previous E1; Dependent on B1 and previous E1 |
| **[8]** | | |
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