Summation: Compute the exact sums

• Feb 9th 2010, 09:15 AM
Runty
Summation: Compute the exact sums
This one's a discrete mathematics question strictly due to its notation.

Compute the exact sums of the following: (SHOW YOUR WORK)

a) $\sum_{i=k}^n7$ where $k$ is an integer between 1 and $n$ (I have no idea how to answer this one)

b) $\sum_{k=1}^n(3k^2-2n+8)$ (I believe I have the answer, but I lack the steps and process to reach the answer)

c) $\sum_{j=1}^n\sum_{k=1}^j\sum_{i=k}^n(i-k)$ (I may have answered this one, but I'd like confirmation)

d) $\sum_{j=18}^{99}12$ (This one was easy, and I've already answered it)

This shouldn't be too hard for some people, but I'd rather not get these wrong.
• Feb 9th 2010, 09:32 AM
Plato
Quote:

Originally Posted by Runty
a) $\sum_{i=k}^n7$ where $k$ is an integer between 1 and $n$ (I have no idea how to answer this one)

$1 \leqslant k \leqslant n \Rightarrow \quad \sum\limits_{i = k}^n 7 = 7\left( {n - k + 1} \right)$
• Feb 9th 2010, 10:03 AM
bmp05
Hi runty,
for 2 you need to rewrite the sum and isolate the term which is being summed:

$
\sum_{k = 1}^n (3k^2 - 2n + 8) = (\sum_{k = 1}^n 3k^2) - n(2n + 8) = 3(\sum_{k = 1}^n k^2) - n(2n + 8)
$

Check this! Do you need $n(2n + 8)$?
Then you can use a closed form of the sum of squares to replace the sum with a relation based on n:

$
\sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}
$

Square pyramidal number - Wikipedia, the free encyclopedia
• Feb 9th 2010, 10:04 AM
danielomalmsteen
b)

$\sum_{k=1}^{n} 3k^2-2n+8 = 3\sum_{k=1}^{n}k^2 -2\sum_{k=1}^{n}n + \sum_{k=1}^{n}8$

$\frac{n(n+1)(2n+1)}{2} - 2n^2 + 8n$