Use standard results from the List of formulae (MF19) to find $\sum_{r=1}^{n} (1 - r - r^2)$ in terms of $n$, simplifying your answer. (b) Show that \begin{equation} \frac{1 - r - r^2}{\left(r^2 + 2r + 2\right)(r + 1)} = \frac{r + 1}{\left(r + 1\right)^2 + 1} - \frac{r}{r^2 + 1} \end{equation} and hence use the method of differences to find $\sum_{r=1}^{n} \frac{1 - r - r^2}{\left(r^2 + 2r + 2\right)(r + 1)}$

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Use standard results from the List of formulae (MF19) to find $\sum_{r=1}^{n} (1 - r - r^2)$ in terms of $n$, simplifying your answer - CIE - A-Level Further Maths - Question 2 - 2021 - Paper 1

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Use standard results from the List of formulae (MF19) to find $\sum_{r=1}^{n} (1 - r - r^2)$ in terms of $n$, simplifying your answer. (b) Show that \begin{equation... show full transcript

Worked Solution & Example Answer:Use standard results from the List of formulae (MF19) to find $\sum_{r=1}^{n} (1 - r - r^2)$ in terms of $n$, simplifying your answer - CIE - A-Level Further Maths - Question 2 - 2021 - Paper 1

Step 1

Find $\sum_{r=1}^{n} (1 - r - r^2)$

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Answer

To find the sum $\sum_{r=1}^{n} (1 - r - r^2)$ , we can separate the terms:

$\sum_{r=1}^{n} (1 - r - r^2) = \sum_{r=1}^{n} 1 - \sum_{r=1}^{n} r - \sum_{r=1}^{n} r^2$

We know that:

$\sum_{r=1}^{n} 1 = n$
$\sum_{r=1}^{n} r = \frac{n(n + 1)}{2}$
$\sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6}$

Combining these results gives:

$\sum_{r=1}^{n} (1 - r - r^2) = n - \frac{n(n + 1)}{2} - \frac{n(n + 1)(2n + 1)}{6}$

This can be simplified to yield the final result in terms of $n$ .

Step 2

Show that$ \frac{1 - r - r^2}{(r^2 + 2r + 2)(r + 1)} = \frac{r + 1}{(r + 1)^2 + 1} - \frac{r}{r^2 + 1}$

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Answer

To show this identity, we can start with the left-hand side:

$\frac{1 - r - r^2}{(r^2 + 2r + 2)(r + 1)}$

Factoring the numerator gives:

$1 - r - r^2 = -(r^2 + r - 1)$

Next, we can manipulate the right-hand side:

$\frac{r + 1}{(r + 1)^2 + 1} - \frac{r}{r^2 + 1}$

Finding a common denominator for the two fractions, we can equate both sides to validate the equality.

Step 3

and hence use the method of differences to find $\sum_{r=1}^{n} \frac{1 - r - r^2}{(r^2 + 2r + 2)(r + 1)}$

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Answer

Using the result from part (b), we can then compute:

$\sum_{r=1}^{n} \left( \frac{r + 1}{(r + 1)^2 + 1} - \frac{r}{r^2 + 1} \right)$

This can be evaluated using the method of differences, recognizing that many terms will cancel out in the series. Ultimately, we can write down the result as:

$\text{Final result} = \text{Value of last term} - \text{Value of first term}$

Calculating these last values will provide us with the total sum.

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Use standard results from the List of formulae (MF19) to find $\sum_{r=1}^{n} (1 - r - r^2)$ in terms of $n$, simplifying your answer - CIE - A-Level Further Maths - Question 2 - 2021 - Paper 1

Worked Solution & Example Answer:Use standard results from the List of formulae (MF19) to find $\sum_{r=1}^{n} (1 - r - r^2)$ in terms of $n$, simplifying your answer - CIE - A-Level Further Maths - Question 2 - 2021 - Paper 1

Find $\sum_{r=1}^{n} (1 - r - r^2)$

Show that\( \frac{1 - r - r^2}{(r^2 + 2r + 2)(r + 1)} = \frac{r + 1}{(r + 1)^2 + 1} - \frac{r}{r^2 + 1}\)

and hence use the method of differences to find $\sum_{r=1}^{n} \frac{1 - r - r^2}{(r^2 + 2r + 2)(r + 1)}$

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