Product and sigma notation

[Rano1234]

The work --- so i understand how to do a & b and i understand the concept of both product and sigma notations, however i have no idea how to use them together. any help would be appreciated

 

[Deleted member 4993]

The work --- so i understand how to do a & b and i understand the concept of both product and sigma notations, however i have no idea how to use them together. any help would be appreciatedView attachment 14620

Please show us - How would you solve:
\(\displaystyle \sum_{i=3}^{5}\frac{j}{i}\) = ?

 

[Rano1234]

Well you'd end up with j/3+j/4+j/5

 

[Deleted member 4993]

Well you'd end up with j/3+j/4+j/5

j/3+j/4+j/5 = 47/60 *j

Now for problem (d) apply product notation to [47/60 *j]

 

[HallsofIvy]

The point is, you don't "apply them together", you apply them "one at a time"!

(c)\(\displaystyle \Sigma_{j= 3}^5\Pi_{i= 4}^5\frac{j}{i}= \Sigma_{j= 3}^5\left(\Pi_{i= 4}^5\frac{j}{i}\right)= \Sigma_{j= 3}^5\left(\frac{j}{4}\right)\left(\frac{j}{5}\right)= \left(\frac{3}{4}\right)\left(\frac{3}{5}\right)+ \left(\frac{4}{4}\right)\left(\frac{4}{5}\right)+ \left(\frac{5}{4}\right)\left(\frac{5}{5}\right)\).

 

[JeffM]

To reiterate Hall's point

[MATH]\sum \prod[/MATH] means the sum of the products.

[MATH]\prod \sum[/MATH] means the product of the sums.

You do the operation on the right and then the operation on the left.

Mechanically, I would do it a bit differently because I hate fractions.

[MATH]\left \{ \sum_{j=3}^5 \left ( \prod_{i=4}^5 \dfrac{j}{i} \right ) \right \} = \left \{ \sum_{j=3}^5 \left ( \dfrac{j}{4} * \dfrac{j}{5} \right ) \right \} = \sum_{j=3}^5 \dfrac{j^2}{20} =[/MATH]
[MATH]\dfrac{1}{20} * \sum_{j=3}^5 j^2 = \dfrac{1}{20} * (3^2 + 4^2 + 5^2) = \dfrac{1}{20} * (9 + 16 + 25) = \dfrac{50}{20} = 2.5.[/MATH]

 

[Rano1234]

thanks for the help guys, i understand how to do these questions now