Moderate -0.5 This is a standard textbook induction proof of a summation formula with a straightforward algebraic verification step. While it requires proper induction structure and careful algebra when expanding (k+1)^3 and (k+1)^2 terms, it follows a completely routine template with no novel insight required. Slightly easier than average due to its formulaic nature, though the algebra is more involved than the simplest induction proofs.
For \(k=1\): \(1^3 = 1\) and \(\frac{1}{4}1^2(1+1)^2 = 1\), so true for \(k=1\)
B1
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Assume true for \(n = k\)
B1
Assuming true for \(k\), \((k+1)\)th term. For alternative statement, give this mark if whole argument logically correct
Next term is \((k+1)^3\); add to both sides
B1
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\(\text{RHS} = \frac{1}{4}k^2(k+1)^2 + (k+1)^3\)
M1
Add to both sides
\(= \frac{1}{4}(k+1)^2\left[k^2 + 4(k+1)\right]\)
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\(= \frac{1}{4}(k+1)^2(k+2)^2\)
M1
Factor of \((k+1)^2\). Allow alternative correct methods
\(= \frac{1}{4}(k+1)^2((k+1)+1)^2\)
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But this is the given result with \((k+1)\) replacing \(k\). Therefore if true for \(k\) it is true for \((k+1)\). Since true for \(k=1\) it is true for \(k = 1, 2, 3, \ldots\)
A1, E1 [7]
A1 for fully convincing algebra leading to true for \(k \Rightarrow\) true for \(k+1\). Accept 'Therefore true by induction' only if previous A1 awarded. S.C. Give E1 if convincing explanation of induction following acknowledgement of earlier error
## Question 6:
For $k=1$: $1^3 = 1$ and $\frac{1}{4}1^2(1+1)^2 = 1$, so true for $k=1$ | B1 | —
Assume true for $n = k$ | B1 | Assuming true for $k$, $(k+1)$th term. For alternative statement, give this mark if whole argument logically correct
Next term is $(k+1)^3$; add to both sides | B1 | —
$\text{RHS} = \frac{1}{4}k^2(k+1)^2 + (k+1)^3$ | M1 | Add to both sides
$= \frac{1}{4}(k+1)^2\left[k^2 + 4(k+1)\right]$ | — | —
$= \frac{1}{4}(k+1)^2(k+2)^2$ | M1 | Factor of $(k+1)^2$. Allow alternative correct methods
$= \frac{1}{4}(k+1)^2((k+1)+1)^2$ | — | —
But this is the given result with $(k+1)$ replacing $k$. Therefore if true for $k$ it is true for $(k+1)$. Since true for $k=1$ it is true for $k = 1, 2, 3, \ldots$ | A1, E1 **[7]** | A1 for fully convincing algebra leading to true for $k \Rightarrow$ true for $k+1$. Accept 'Therefore true by induction' only if previous A1 awarded. S.C. Give E1 if convincing explanation of induction following acknowledgement of earlier error
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