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Paper 1 · November 2010 · Series and Proof by Induction

Given that $\sum_{r=k+1}^{3k} r^3 = k^2(4k+1)(5k+2)$ for $k \geq 1$ , evaluate $21^3 + 22^3 + 23^3 + \ldots + 60^3$ .

Model answer

3304800

Also accepted: 3 304 800, 3,304,800

Explanation

The first term is

21^3

, so

k+1=21

, and the last is

60^3

, so

3k=60

. Both give

k=20

. Then

k^2(4k+1)(5k+2)=400\times81\times102=400\times8262=3\,304\,800

.

Derived from ZIMSEC Further Mathematics 9187/1, November 2010, Q3

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More questions from this paper

The set of positive rational numbers forms a group under the operation $a * b = \frac{ab}{10}$ . Find the identity element.
Under the operation $a * b = \frac{ab}{10}$ on the positive rational numbers, write the inverse of an element $a$ in terms of $a$ .
The positive rational numbers form a group under $a * b = \frac{ab}{10}$ . Which group axiom fails if the set is narrowed to the positive integers?
The set $G = \{0, 1, 2\}$ forms a group under addition modulo 3. State the identity element.
In the group $G = \{0, 1, 2\}$ under addition modulo 3, find the inverse of the element 2.

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