Challenging +1.2 This is a structured multi-part induction question with standard techniques. The initial induction proof of the sum of cubes formula is routine for Further Maths students. The subsequent parts involve algebraic manipulation using known summation formulas, finding a constant to achieve a desired form, and proving inequalities—all guided and methodical. While it requires careful algebra across multiple steps, it doesn't demand novel insight or particularly challenging problem-solving beyond FP1 expectations.
Prove by induction that
$$\sum _ { n = 1 } ^ { N } n ^ { 3 } = \frac { 1 } { 4 } N ^ { 2 } ( N + 1 ) ^ { 2 }$$
Use this result, together with the formula for \(\sum _ { n = 1 } ^ { N } n ^ { 2 }\), to show that
$$\sum _ { n = 1 } ^ { N } \left( 20 n ^ { 3 } + 36 n ^ { 2 } \right) = N ( N + 1 ) ( N + 3 ) ( 5 N + 2 ) .$$
Let
$$S _ { N } = \sum _ { n = 1 } ^ { N } \left( 20 n ^ { 3 } + 36 n ^ { 2 } + \mu n \right)$$
Find the value of the constant \(\mu\) such that \(S _ { N }\) is of the form \(N ^ { 2 } ( N + 1 ) ( a N + b )\), where the constants \(a\) and \(b\) are to be determined.
Show that, for this value of \(\mu\),
$$5 + \frac { 22 } { N } < N ^ { - 4 } S _ { N } < 5 + \frac { 23 } { N }$$
for all \(N \geqslant 18\).
Prove by induction that
$$\sum _ { n = 1 } ^ { N } n ^ { 3 } = \frac { 1 } { 4 } N ^ { 2 } ( N + 1 ) ^ { 2 }$$
Use this result, together with the formula for $\sum _ { n = 1 } ^ { N } n ^ { 2 }$, to show that
$$\sum _ { n = 1 } ^ { N } \left( 20 n ^ { 3 } + 36 n ^ { 2 } \right) = N ( N + 1 ) ( N + 3 ) ( 5 N + 2 ) .$$
Let
$$S _ { N } = \sum _ { n = 1 } ^ { N } \left( 20 n ^ { 3 } + 36 n ^ { 2 } + \mu n \right)$$
Find the value of the constant $\mu$ such that $S _ { N }$ is of the form $N ^ { 2 } ( N + 1 ) ( a N + b )$, where the constants $a$ and $b$ are to be determined.
Show that, for this value of $\mu$,
$$5 + \frac { 22 } { N } < N ^ { - 4 } S _ { N } < 5 + \frac { 23 } { N }$$
for all $N \geqslant 18$.
\hfill \mbox{\textit{CAIE FP1 2009 Q11 EITHER}}