3. (a) Show that \(r ^ { 3 } - ( r - 1 ) ^ { 3 } \equiv 3 r ^ { 2 } - 3 r + 1\)
(b) Hence prove by the method of differences that, for \(n \in \mathbb { Z } ^ { + }\)
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }$$
[You may use \(\sum _ { r = 1 } ^ { n } r = \frac { n ( n + 1 ) } { 2 }\) without proof.]
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Question 3:
Part (a):
Answer Marks
Guidance
Answer/Working Mark
Guidance Notes
\(r^3 - (r-1)^3 \equiv r^3 - (r^3 - 3r^2 + 3r - 1) \equiv 3r^2 - 3r + 1\) B1*
Shows correct expansion of \((r-1)^3\) or uses \(a^3-b^3=(a-b)(a^2+ab+b^2)\) and achieves printed answer with no errors
Part (b):
Answer Marks
Guidance
Answer/Working Mark
Guidance Notes
Differences method: \(n^3 - (n-1)^3\), \((n-1)^3-(n-2)^3\), ..., \(1^3-0^3\) M1
Uses method of differences. Must include at least \(r=1,2,...,n\) or \(r=1,...,(n-1),n\)
\(n^3 = \sum_{r=1}^{n}(3r^2-3r+1) = \sum_{r=1}^{n}3r^2 - \sum_{r=1}^{n}3r + \sum_{r=1}^{n}1\) M1
Sets \(n^3 = \sum_{r=1}^{n}(3r^2-3r+1)\) and attempts to expand RHS
\(\sum_{r=1}^{n}1 = n\) B1
\(\sum_{r=1}^{n}1 = n\) seen or implied
\(3\sum_{r=1}^{n}r^2 = n(n-1)(n+1) + \frac{3}{2}n(n+1)\) dM1
Rearranges to make \(k\sum_{r=1}^{n}r^2\) the subject and substitutes for \(\sum_{r=1}^{n}r\). Dependent on first M mark
\(\sum_{r=1}^{n}r^2 = \frac{1}{6}n(n+1)(2n+1)\) A1*
Completely correct solution with no errors seen
[Total: 6]
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# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $r^3 - (r-1)^3 \equiv r^3 - (r^3 - 3r^2 + 3r - 1) \equiv 3r^2 - 3r + 1$ | B1* | Shows correct expansion of $(r-1)^3$ or uses $a^3-b^3=(a-b)(a^2+ab+b^2)$ and achieves printed answer with no errors |
## Part (b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Differences method: $n^3 - (n-1)^3$, $(n-1)^3-(n-2)^3$, ..., $1^3-0^3$ | M1 | Uses method of differences. Must include at least $r=1,2,...,n$ or $r=1,...,(n-1),n$ |
| $n^3 = \sum_{r=1}^{n}(3r^2-3r+1) = \sum_{r=1}^{n}3r^2 - \sum_{r=1}^{n}3r + \sum_{r=1}^{n}1$ | M1 | Sets $n^3 = \sum_{r=1}^{n}(3r^2-3r+1)$ and attempts to expand RHS |
| $\sum_{r=1}^{n}1 = n$ | B1 | $\sum_{r=1}^{n}1 = n$ seen or implied |
| $3\sum_{r=1}^{n}r^2 = n(n-1)(n+1) + \frac{3}{2}n(n+1)$ | dM1 | Rearranges to make $k\sum_{r=1}^{n}r^2$ the subject and substitutes for $\sum_{r=1}^{n}r$. **Dependent on first M mark** |
| $\sum_{r=1}^{n}r^2 = \frac{1}{6}n(n+1)(2n+1)$ | A1* | Completely correct solution with no errors seen |
**[Total: 6]**
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3. (a) Show that $r ^ { 3 } - ( r - 1 ) ^ { 3 } \equiv 3 r ^ { 2 } - 3 r + 1$\\
(b) Hence prove by the method of differences that, for $n \in \mathbb { Z } ^ { + }$
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }$$
[You may use $\sum _ { r = 1 } ^ { n } r = \frac { n ( n + 1 ) } { 2 }$ without proof.]\\
\hfill \mbox{\textit{Edexcel F2 2017 Q3 [6]}}