sum of series using method of difference Please I have a problem with finding d sum of the sequence 3x4 ,4x5 ,5x6,...... using method of difference ....most books I use only explain partial fractions,  but I have found the $n$th term to be $(n + 2) (n +3)$    ....how can I show that the sum is $n(n^2 + 9n + 26) / 3$?
I tried
$$\begin{array}{lc}
   r = 1    & (4)(5)   - (3)(4)\\
   r = 2  &   (5)(6)   - (4)(5) \\
  r = n-1 & (n+2)(n+3)    -  (n + 1)(n+2)\\
r= n   & (n+3)(n+4)  - (n+2)(n+3)
\end{array}$$
and got $(n^2 + 7n +12) / 2       -      (3)(4)$     but I don't know if that's the way.


*

*pls can a good book be suggested for me?

 A: $a_n = n$ has a first difference of $a_n-a_{n-1} = n-(n-1) = 1$. 
$b_n = n^2$ has a first difference of $b_n-b_{n-1} = n^2-(n-1)^2 = 2n-1$. 
$c_n = n^3$ has a first difference of $c_n-c_{n-1} = n^3-(n-1)^3 = 3n^2-3n+1$.
$d_n = \displaystyle\sum_{k=1}^{n}(k+2)(k+3)$ has a first difference of $d_n-d_{n-1} = (n+2)(n+3) = n^2+5n+6$. 
Also, $d_0 = 0$. Based on this information, can you write $d_n$ as a linear combination of $a_n$, $b_n$, $c_n$?
A: the difference between consecutive products of $k$ consecutive integers is a constant times a product of $k-1$ consecutive integers:
the case $k=1$ is trivial:
$$
(n+1) - n = 1
$$
then we have:
$$
(n+2)(n+1) - (n+1)n = 2(n+1) \\
(n+3)(n+2)(n+1) - (n+2)(n+1)n = 3(n+2)(n+1) \\
(n+4)(n+3)(n+2)(n+1) - (n+3)(n+2)(n+1)n = 4(n+3)(n+2)(n+1)\\
\cdots
$$
these relations may be used to form telescoping sums. so, for example,
$$
3.4 + 4.5 + 5.6 + \cdots +(n+1)(n+2) = \sum_{k=2}^n (k+1)(k+2) \\
= \frac13 \sum_{k=2}^n \left[(k+3)(k+2)(k+1) - (k+2)(k+1)k\right] \\
= \frac13 \left[(n+3)(n+2)(n+1) - 24 \right]
$$ 
