I've never come across a series like this, not sure how to sum it:
$\displaystyle
1.2.6 + 2.3.7 + ... + n(n+1)(n+5)
$
Maybe something to do with factorials? No idea :/
One way is........
$\displaystyle \displaystyle\sum_{k=1}^nk(k+1)(k+5)=\sum_{k=1}^n\ left[k^3+6k^2+5k\right]$
The sum of cubes is
$\displaystyle \displaystyle\ 1^3+2^3+3^3+....+n^3=(1+2+....+n)^2=\left[\frac{n(n+1)}{2}\right]^2$
The formulae for the "sum of n squares" and the "sum of n terms" can be used for the other two parts.
I took it straight from the exam paper: http://mathsathawthorn.pbworks.com/f/FP1Jun06.pdf
Oh I see, that's kinda of easy really. Thought there'd be some special trickery or something.
Thanks
In some cultures and age groups, the multiplicative dot and decimal point are used the other way around.
E.g. in Australia, most older generations use a centred dot for a decimal point and a full stop for multiplication, while younger (e.g. mine) generations use the centred dot for multiplication and the full stop for decimals. I assume this is from the introduction of computers - decimals would be used much more frequently than multiplicative dots, so it's easier to use the full stop key for a decimal point.
If you have no formulae to work with, you could analyse patterns...
$\displaystyle k=1,\;2,\;3,\;4,\;$
$\displaystyle \sum\;k=1,\;3,\;6,\;10,....$
$\displaystyle k^3=1,\;8,\;27,\;64,....$
$\displaystyle \sum\;k^3=1,\;9,\;36,\;100,....$
From the pattern, it seems
$\displaystyle \sum\;k^3=\left(\sum\;k\right)^2$
To prove that beyond doubt, you can apply "Proof By Induction".
Similarly, you can arrive at a formula for the sum of squares...
$\displaystyle k=1,\;2,\;3,\;4,.....$
$\displaystyle \sum\;k=1,\;3,\;6,\;10,..$
$\displaystyle k^2=1,\;4,\;9,\;16,.....$
$\displaystyle \sum\;k^2=1,\;5,\;14,\;30,...$
$\displaystyle \displaystyle\frac{\sum\;k^2}{\sum\;k}=\frac{1}{1} ,\;\frac{5}{3},\;\frac{14}{6},\;\frac{30}{10},.... =\frac{3}{3},\;\frac{5}{3},\;\frac{7}{3},\;\frac{9 }{3},....$
which is an arithmetic sequence pattern of
$\displaystyle \displaystyle\ U_n=a+(n-1)d=1+(n-1)\frac{2}{3}=\frac{3+2n-2}{3}=\frac{2n+1}{3}$
Hence, it seems
$\displaystyle \displaystyle\frac{\sum\;k^2}{\sum\;k}=\frac{2n+1} {3}\Rightarrow\sum\;k^2=\left[\frac{2n+1}{3}\right]\sum\;k$
Proof By Induction can also verify this...
Hence all 3 sums can be expressed in terms of $\displaystyle \displaystyle\left[\frac{n(n+1)}{2}\right]$
We seem to use both here,
$\displaystyle
5 \cdot 7, \sin\theta.t, 0.5634
$
If I'm writing it I tend to make multiplication dots thicker.
I'm still trying to digest this ha.
It looks interesting but may be straying too far from the default answers, you're not expected to be at all intuitive in your answers for these papers.
I get
$\displaystyle \displaystyle\left[\frac{n(n+1)}{2}\right]^2+\frac{6n(n+1)}{2}\;\frac{(2n+1)}{3}+\frac{5n(n+ 1)}{2}$
$\displaystyle =\displaystyle\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+2(2n+1)+5\right]=\frac{n(n+1)}{2}\left[\frac{n^2+n+8n+4+10}{2}\right]$
$\displaystyle =\displaystyle\frac{n(n+1)}{2}\left[\frac{n^2+9n+14}{2}\right]=\frac{n(n+1)}{2}\frac{(n+2)(n+7)}{2}$