Challenging +1.2 This is a Further Maths question requiring application of Euler's approximation formula to verify a known summation result. While it involves Further Maths content and requires careful algebraic manipulation through multiple steps, the question is essentially verification rather than discovery—students apply a given formula to a standard sum and show it produces the expected result. The algebraic work is substantial but methodical, making it moderately challenging but not requiring novel insight.
13 Prove that Euler's approximate formula, as given in line 13, when applied to \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \mathrm { r } ^ { 2 }\) gives exactly \(\frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }\).
AO 2.2a. Correct completion. Intermediate step needed. Dep on B3. OR convincing comparison of \(\frac{n(n+1)(2n+1)}{6}\) and \(\frac{n^3-1}{3}+\frac{6n^2+2n+4}{12}\)
## Question 13:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_1^n x^2\,dx + \frac{n^2+1^2}{2} + \frac{1^2-2^2}{12} - \frac{n^2-(n+1)^2}{12}$ | B1 | AO 1.1a. Substitute correctly into formula. Condone $r$ but not $n$ instead of $x$ |
| $\left[\frac{x^3}{3}\right]_1^n + \frac{6n^2+6+1-4-n^2+n^2+2n+1}{12}$ | | |
| $\frac{n^3-1}{3} + \left[\frac{6n^2+2n+4}{12}\right]$ | B1 | AO 1.1. Integral correctly evaluated |
| $\frac{2n^3+3n^2+n}{6}$ | B1 | AO 2.1. Correct single fraction – may not be fully simplified. OR $\frac{n(n+1)(2n+1)}{6} = \frac{2n^3+3n^2+n}{6}$ |
| $\frac{n(2n^2+3n+1)}{6} = \frac{n(n+1)(2n+1)}{6}$ | B1 [4] | AO 2.2a. Correct completion. Intermediate step needed. Dep on B3. OR convincing comparison of $\frac{n(n+1)(2n+1)}{6}$ and $\frac{n^3-1}{3}+\frac{6n^2+2n+4}{12}$ |
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13 Prove that Euler's approximate formula, as given in line 13, when applied to $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \mathrm { r } ^ { 2 }$ gives exactly $\frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }$.
\hfill \mbox{\textit{OCR MEI Paper 3 2023 Q13 [4]}}