Answers 1 9 r − 3 − 1 9 r + 6 . {\frac{ 1 }{ 9 r - 3 } - \frac{ 1 }{ 9 r + 6 }.} 9 r − 3 1 − 9 r + 6 1 . 2 m + 1 ( 9 m + 2 ) ( 3 m − 1 ) . {\frac{2 m + 1}{(9 m + 2)(3 m - 1)}.} ( 9 m + 2 ) ( 3 m − 1 ) 2 m + 1 . Smallest
n = 28. {n = 28.} n = 28. Full solutions
(a) 1 9 r 2 + 3 r − 2 = 1 ( 3 r − 1 ) ( 3 r + 2 ) 1 ( 3 r − 1 ) ( 3 r + 2 ) = A 3 r − 1 + B 3 r + 2 \begin{align*}
\frac{1}{9 r^2 + 3 r - 2} &= \frac{1}{(3 r - 1)(3 r + 2)} \\
\frac{1}{(3 r - 1)(3 r + 2)} &= \frac{A}{3 r - 1} + \frac{B}{3 r + 2}
\end{align*} 9 r 2 + 3 r − 2 1 ( 3 r − 1 ) ( 3 r + 2 ) 1 = ( 3 r − 1 ) ( 3 r + 2 ) 1 = 3 r − 1 A + 3 r + 2 B
By the cover-up rule,
A = 1 3 ( 1 3 ) + 2 = 1 3 B = 1 3 ( − 2 3 ) − 1 = − 1 3 \begin{align*}
A &= \frac{1}{3\left(\frac{1}{3}\right) + 2}
\\ &= \frac{1}{3}
\\ B &= \frac{1}{3\left(- \frac{2}{3}\right) - 1}
\\ &= - \frac{1}{3}
\end{align*} A B = 3 ( 3 1 ) + 2 1 = 3 1 = 3 ( − 3 2 ) − 1 1 = − 3 1
Hence
1 9 r 2 + 3 r − 2 = 1 3 ( 3 r − 1 ) − 1 3 ( 3 r + 2 ) = 1 9 r − 3 − 1 9 r + 6 ■ \begin{align*}
\frac{1}{9 r^2 + 3 r - 2} &= \frac{1}{3(3 r - 1)} - \frac{1}{3(3 r + 2)}
\\ &= \frac{ 1 }{ 9 r - 3 } - \frac{ 1 }{ 9 r + 6 } \; \blacksquare
\end{align*} 9 r 2 + 3 r − 2 1 = 3 ( 3 r − 1 ) 1 − 3 ( 3 r + 2 ) 1 = 9 r − 3 1 − 9 r + 6 1 ■
(b) ∑ r = m 3 m 1 9 r 2 + 3 r − 2 = ∑ r = m 3 m ( 1 9 r − 3 − 1 9 r + 6 ) = 1 9 m − 3 − 1 9 m + 6 + 1 9 m + 6 − 1 9 m + 15 + 1 9 m + 15 − 1 9 m + 24 + ⋯ + 1 27 m − 21 − 1 27 m − 12 + 1 27 m − 12 − 1 27 m − 3 + 1 27 m − 3 − 1 27 m + 6 = 1 9 m − 3 − 1 27 m + 6 = 27 m + 6 − ( 9 m − 3 ) 9 ( 3 m − 1 ) ( 9 m + 2 ) = 2 m + 1 ( 3 m − 1 ) ( 9 m + 2 ) ■ \begin{align*}
& \sum_{r=m}^{3m} \frac{1}{9 r^2 + 3 r - 2} \\
& = \sum_{r=m}^{3m} \left( \frac{ 1 }{ 9 r - 3 } - \frac{ 1 }{ 9 r + 6 } \right) \\
& = \def\arraystretch{1.5}
\begin{array}{lclc}
& \frac{ 1 }{ 9 m - 3 } &-& \cancel{\frac{ 1 }{ 9 m + 6 }} \\
+ & \cancel{\frac{ 1 }{ 9 m + 6 }} &-& \cancel{\frac{ 1 }{ 9 m + 15 }} \\
+ & \cancel{\frac{ 1 }{ 9 m + 15 }} &-& \cancel{\frac{ 1 }{ 9 m + 24 }} \\
+ & & \cdots & \\
+ & \cancel{\frac{ 1 }{ 27 m - 21 }} &-& \cancel{\frac{ 1 }{ 27 m - 12 }} \\
+ & \cancel{\frac{ 1 }{ 27 m - 12 }} &-& \cancel{\frac{ 1 }{ 27 m - 3 }} \\
+ & \cancel{\frac{ 1 }{ 27 m - 3 }} &-& \frac{ 1 }{ 27 m + 6 } \\
\end{array} \\
&= \frac{ 1 }{ 9 m - 3 } - \frac{ 1 }{ 27 m + 6 } \\
&= \frac{27m+6 - (9m-3)}{9(3m-1)(9m+2)} \\
&= \frac{2m+1}{(3m-1)(9m+2)} \; \blacksquare
\end{align*} r = m ∑ 3 m 9 r 2 + 3 r − 2 1 = r = m ∑ 3 m ( 9 r − 3 1 − 9 r + 6 1 ) = + + + + + + 9 m − 3 1 9 m + 6 1 9 m + 15 1 27 m − 21 1 27 m − 12 1 27 m − 3 1 − − − ⋯ − − − 9 m + 6 1 9 m + 15 1 9 m + 24 1 27 m − 12 1 27 m − 3 1 27 m + 6 1 = 9 m − 3 1 − 27 m + 6 1 = 9 ( 3 m − 1 ) ( 9 m + 2 ) 27 m + 6 − ( 9 m − 3 ) = ( 3 m − 1 ) ( 9 m + 2 ) 2 m + 1 ■
(c) ∑ r = 1 n 1 9 r 2 + 3 r − 2 = ∑ r = 1 n ( 1 9 r − 3 − 1 9 r + 6 ) = 1 6 − 1 15 1 15 − 1 24 + ⋯ + 1 9 n − 12 − 1 9 n − 3 + 1 9 n − 3 − 1 9 n + 6 = 1 6 − 1 9 n + 6 \begin{align*}
& \sum_{r=1}^{n} \frac{1}{9 r^2 + 3 r - 2} \\
& = \sum_{r=1}^{n} \left( \frac{ 1 }{ 9 r - 3 } - \frac{ 1 }{ 9 r + 6 } \right) \\
& = \def\arraystretch{1.5}
\begin{array}{lclc}
& \frac{1}{6} &-& \cancel{\frac{1}{15}} \\
& \cancel{\frac{1}{15}} &-& \cancel{\frac{1}{24}} \\
+ & & \cdots & \\
+ & \cancel{\frac{ 1 }{ 9 n - 12 }} &-& \cancel{\frac{ 1 }{ 9 n - 3 }} \\
+ & \cancel{\frac{ 1 }{ 9 n - 3 }} &-& \frac{ 1 }{ 9 n + 6 } \\
\end{array} \\
&= \frac{1}{6} - \frac{ 1 }{ 9 n + 6 }
\end{align*} r = 1 ∑ n 9 r 2 + 3 r − 2 1 = r = 1 ∑ n ( 9 r − 3 1 − 9 r + 6 1 ) = + + + 6 1 15 1 9 n − 12 1 9 n − 3 1 − − ⋯ − − 15 1 24 1 9 n − 3 1 9 n + 6 1 = 6 1 − 9 n + 6 1
As
n → ∞ , {n \to \infty,} n → ∞ , − 1 9 n + 6 → ∞ . {\frac{ - 1 }{ 9 n + 6 } \to \infty.} 9 n + 6 − 1 → ∞. ∑ r = 1 ∞ 1 9 r 2 + 3 r − 2 = 1 6 ■ \sum_{r=1}^\infty \frac{1}{9 r^2 + 3 r - 2} = \frac{1}{6} \; \blacksquare r = 1 ∑ ∞ 9 r 2 + 3 r − 2 1 = 6 1 ■
(d) ∣ ∑ r = 1 ∞ ( 1 9 r 2 + 3 r − 2 ) − ∑ r = 1 n ( 1 9 r 2 + 3 r − 2 ) ∣ < 0.004 ∣ 1 6 − ( 1 6 − 1 9 n + 6 ) ∣ < 0.004 1 9 n + 6 < 1 250 \begin{gather*}
\left| \sum_{r=1}^\infty \left( \frac{1}{9 r^2 + 3 r - 2} \right) - \sum_{r=1}^n \left( \frac{1}{9 r^2 + 3 r - 2} \right) \right| < 0.004
\\ \left| \frac{1}{6} - \left( \frac{1}{6} - \frac{ 1 }{ 9 n + 6 } \right) \right| < 0.004
\\ \frac{1}{9n+6} < \frac{1}{250}
\end{gather*} r = 1 ∑ ∞ ( 9 r 2 + 3 r − 2 1 ) − r = 1 ∑ n ( 9 r 2 + 3 r − 2 1 ) < 0.004 6 1 − ( 6 1 − 9 n + 6 1 ) < 0.004 9 n + 6 1 < 250 1
Since
9 n + 6 > 0 , {9n+6 > 0,} 9 n + 6 > 0 , 9 n + 6 > 250 n > 244 9 \begin{align*}
9 n + 6 &> 250
\\ n &> \frac{244}{9}
\end{align*} 9 n + 6 n > 250 > 9 244 Smallest n = 28 ■ \textrm{Smallest } n = 28 \; \blacksquare Smallest n = 28 ■ Question Commentary The method of differences approach via partial fractions came out
again (having appeared in 2021 Paper 1 Question 5
and 2019 Paper 1 Question 6 ).
I find the 1 mark allocated for part (c) interesting though. While not tough as we
just have to do the same method of differences as in part (b) (and it's even easier to
manage with the more common r = 1 {r=1} r = 1 n {n} n
I have tried to look at others' attempt at this but have not found a drastically more elegant
alternative yet.