Standard +0.3 This is a straightforward FP1 question requiring direct application of standard summation formulae. Part (a) involves algebraic manipulation of known results (routine for Further Maths students), and part (b) requires substituting n=12 into the result from (a) and solving for k using the geometric series formula. No novel insight or complex problem-solving is needed—just methodical application of standard techniques.
8. (a) Use the standard results for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) to show that
$$\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 2 } + 8 r + 3 \right) = \frac { 1 } { 2 } n ( 2 n + 5 ) ( n + 3 )$$
for all positive integers \(n\).
Given that
$$\sum _ { r = 1 } ^ { 12 } \left( 3 r ^ { 2 } + 8 r + 3 + k \left( 2 ^ { r - 1 } \right) \right) = 3520$$
(b) find the exact value of the constant \(k\).
Attempt to apply sum to 12 terms of GP or adds up all 12 terms; \(\frac{1(1-2^{12})}{1-2}\) o.e. or 4095
\(2610 + 4095k = 3520 \Rightarrow k = \frac{2}{9}\)
A1
\(k = \frac{2}{9}\) or \(0.\dot{2}\)
Note for 8(b): \(2^{\text{nd}}\) M1 \(1^{\text{st}}\) A1: These two marks can be implied by seeing 4095 or \(4095k\)
# Question 8:
## Part (a):
| $\sum_{r=1}^{n}(3r^2+8r+3) = \frac{3}{6}n(n+1)(2n+1) + \frac{8}{2}n(n+1) + 3n$ | M1 | Attempt to use at least one correct standard formula for first two terms |
| Correct first two terms | A1 | |
| $3 \to 3n$ | B1 | |
| Factor out at least $n$ from all terms | M1 | There must be a factor of $n$ in every term |
| $= \frac{1}{2}n(2n^2+11n+15) = \frac{1}{2}n(2n+5)(n+3)$ (*) | A1* cso | Achieves correct answer, no errors seen |
## Part (b):
| $\sum_{r=1}^{12}(3r^2+8r+3) = \frac{1}{2}(12)(29)(15) \{= 2610\}$ | M1 | Attempt to evaluate $\sum_{r=1}^{12}(3r^2+8r+3)$ |
| $\sum_{r=1}^{12}(2^{r-1}) = \frac{1(1-2^{12})}{1-2} \{= 4095\}$ | M1; A1 | Attempt to apply sum to 12 terms of GP or adds up all 12 terms; $\frac{1(1-2^{12})}{1-2}$ o.e. or 4095 |
| $2610 + 4095k = 3520 \Rightarrow k = \frac{2}{9}$ | A1 | $k = \frac{2}{9}$ or $0.\dot{2}$ |
**Note for 8(b):** $2^{\text{nd}}$ M1 $1^{\text{st}}$ A1: These two marks can be implied by seeing 4095 or $4095k$
8. (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that
$$\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 2 } + 8 r + 3 \right) = \frac { 1 } { 2 } n ( 2 n + 5 ) ( n + 3 )$$
for all positive integers $n$.
Given that
$$\sum _ { r = 1 } ^ { 12 } \left( 3 r ^ { 2 } + 8 r + 3 + k \left( 2 ^ { r - 1 } \right) \right) = 3520$$
(b) find the exact value of the constant $k$.\\
\hfill \mbox{\textit{Edexcel FP1 2017 Q8 [9]}}