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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?
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$a_n = n$ 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 $a_n$, $b_n$, $c_n$?

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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, $$ 3.4 + 4.5 + 5.6 + \cdots +(n+1)(n+2) = \sum_{k=2}^n (k+1)(k+2) \\ = \frac13 \sum_{k=2}^n \left[(k+3)(k+2)(k+1) - (k+2)(k+1)k\right] \\ = \frac13 \left[(n+3)(n+2)(n+1) - 24 \right] $$

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  • $\begingroup$ thnx to u guys .... it helped me $\endgroup$ – user3572546 Dec 1 '14 at 20:12

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