Standard summation formula application

1. (a) Use the standard results for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that

$$\sum _ { r = 1 } ^ { n } ( 3 r - 2 ) ^ { 2 } = \frac { 1 } { 2 } n \left[ 6 n ^ { 2 } - 3 n - 1 \right]$$ for all positive integers \(n\).
(b) Hence find any values of \(n\) for which $$\sum _ { r = 5 } ^ { n } ( 3 r - 2 ) ^ { 2 } + 103 \sum _ { r = 1 } ^ { 28 } r \cos \left( \frac { r \pi } { 2 } \right) = 3 n ^ { 3 }$$

1. An art display consists of an arrangement of \(n\) marbles.

When arranged in ascending order of mass, the mass of the first marble is 10 grams. The mass of each subsequent marble is 3 grams more than the mass of the previous one, so that the \(r\) th marble has mass \(( 7 + 3 r )\) grams.

1. Show that the mean mass, in grams, of the marbles in the display is given by $$\frac { 1 } { 2 } ( 3 n + 17 )$$ Given that there are 85 marbles in the display,
2. use the standard summation formulae to find the standard deviation of the mass of the marbles in the display, giving your answer, in grams, to one decimal place.

5. \begin{figure}[h] \end{figure} A block has length \(( r + 2 ) \mathrm { cm }\), width \(( r + 1 ) \mathrm { cm }\) and height \(r \mathrm {~cm}\), as shown in Figure 2.
In a set of \(n\) such blocks, the first block has a height of 1 cm , the second block has a height of 2 cm , the third block has a height of 3 cm and so on.

1. Use the standard results for \(\sum _ { r = 1 } ^ { n } r ^ { 3 } , \sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that the total volume, \(V\), of all \(n\) blocks in the set is given by $$V = \frac { n } { 4 } ( n + 1 ) ( n + 2 ) ( n + 3 ) \quad n \geqslant 1$$ Given that the total volume of all \(n\) blocks is $$\left( n ^ { 4 } + 6 n ^ { 3 } - 11710 \right) \mathrm { cm } ^ { 3 }$$
2. determine how many blocks make up the set.

1. (a) Use the standard results for summations to show that for all positive integers \(n\)

$$\sum _ { r = 1 } ^ { n } ( 5 r - 2 ) ^ { 2 } = \frac { 1 } { 6 } n \left( a n ^ { 2 } + b n + c \right)$$ where \(a\), \(b\) and \(c\) are integers to be determined.
(b) Hence determine the value of \(k\) for which $$\sum _ { r = 1 } ^ { k } ( 5 r - 2 ) ^ { 2 } = 94 k ^ { 2 }$$

1. (a) Use the standard summation formulae to show that, for \(n \in \mathbb { N }\),

$$\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 2 } - 17 r - 25 \right) = n \left( n ^ { 2 } - A n - B \right)$$ where \(A\) and \(B\) are integers to be determined.
(b) Explain why, for \(k \in \mathbb { N }\), $$\sum _ { r = 1 } ^ { 3 k } r \tan ( 60 r ) ^ { \circ } = - k \sqrt { 3 }$$ Using the results from part (a) and part (b) and showing all your working,
(c) determine any value of \(n\) that satisfies $$\sum _ { r = 5 } ^ { n } \left( 3 r ^ { 2 } - 17 r - 25 \right) = 15 \left[ \sum _ { r = 6 } ^ { 3 n } r \tan ( 60 r ) ^ { \circ } \right] ^ { 2 }$$

1. (a) Use the standard results for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that, for all positive integers \(n\),

$$\sum _ { r = 1 } ^ { n } ( 2 r - 1 ) ^ { 2 } = \frac { n } { 3 } \left( a n ^ { 2 } - 1 \right)$$ where \(a\) is a constant to be determined.
(b) Hence determine the sum of the squares of all positive odd three-digit integers.

1. (a) Use the standard results for summations to show that, for all positive integers \(n\),

$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( r + 1 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( a n + b )$$ where \(a\) and \(b\) are integers to be determined.
(b) Hence show that, for all positive integers \(k\), $$\sum _ { r = k + 1 } ^ { 3 k } r ^ { 2 } ( r + 1 ) = \frac { 1 } { 3 } k ( 3 k + 1 ) \left( A k ^ { 2 } + B k + C \right)$$ where \(A , B\) and \(C\) are integers to be determined.
(c) Hence, using algebra and making your method clear, determine the value of \(k\) for which $$25 \sum _ { r = k + 1 } ^ { 3 k } r ^ { 2 } ( r + 1 ) = 192 k ^ { 3 } ( 3 k + 1 )$$