1. ## series, sum help

show that
$\displaystyle \displaystyle\sum_{r=1}^n \frac{r+3}{2} = kn(n+7)$
where K is a rational constant to be found

so I just plug in numbers for 'r' e.g. r=1,2,3 , n and get the sequence 2, 5/2 , 3 ,.....$\displaystyle \frac{ n+3}{2}$

but not sure how to prove it equals the expression $\displaystyle kn(n+7)$

2. Hello !

The sum can be rewritten this way :

$\displaystyle \frac 12 \sum_{r=1}^n r +\frac 32\sum_{r=1}^n 1$

The first one is known : $\displaystyle \sum_{r=1}^n r=\frac{n(n+1)}{2}$

And $\displaystyle \sum_{r=1}^n 1=n$ (you add n times 1)

So the sum is $\displaystyle \frac{n(n+1)}{4}+\frac{3n}{2}=\frac{n(n+7)}{4}$

3. $\displaystyle \sum\limits_{r = 1}^n {\frac{{r + 3}} {2}} = \frac{1} {2}\sum\limits_{r = 1}^n {r + 3 = \frac{1} {2}\left[ {\sum\limits_{r = 1}^n r + \sum\limits_{r = 1}^n 3 } \right]}$

$\displaystyle \frac{1} {2}\left[ {\sum\limits_{r = 1}^n r + \sum\limits_{r = 1}^n 3 } \right] = \frac{1} {2}\left[ {\frac{{n(n + 1)}} {2} + 3n} \right] = \frac{1} {2}\left[ {\frac{{n^2 + 7n}} {2}} \right] = \frac{1} {4}n\left( {n + 7} \right)$

4. [quote]
Originally Posted by Moo
Hello !

The sum can be rewritten this way :

$\displaystyle \frac 12 \sum_{r=1}^n r +\frac 32\sum_{r=1}^n 1$
The first one is known : $\displaystyle \sum_{r=1}^n r=\frac{n(n+1)}{2}$

I dont understand it from here, how is this know?

thanks

And $\displaystyle \sum_{r=1}^n 1=n$ (you add n times 1)

So the sum is $\displaystyle \frac{n(n+1)}{4}+\frac{3n}{2}=\frac{n(n+7)}{4}$

5. This is one of the best known and widely used sums: $\displaystyle \sum\limits_{k = 1}^J k = \frac{{J(J + 1)}}{2}$.

If you add J to itself N times you get NJ so: $\displaystyle \sum\limits_{k = 1}^N J = NJ$.

There are several of these special sums that you need to learn.

6. Originally Posted by Plato
This is one of the best known and widely used sums: $\displaystyle \sum\limits_{k = 1}^J k = \frac{{J(J + 1)}}{2}$.

If you add J to itself N times you get NJ so: $\displaystyle \sum\limits_{k = 1}^N J = NJ$.

There are several of these special sums that you need to learn.
Then there must be an easier way of doing this, becasue I have not learnt any of the above stuff mentioned.

could I not just take the first term, 2 and the common difference 1/2 and use the formaul for the sum of an artihmetic series?

$\displaystyle S_{n} = \frac{n}{2}[2a+(n-1)d]$

would that be correct ?

7. Yes, of course you can. But your original post gave no clue that you knew that formula so responders were telling you how to do it without that formula. In fact, that formula is derived by just the method they describe.