# Sum of Products Help

1. Sep 3, 2009

### tstuddud

I don't quite understand the method to solve this type of question.

Let x=(-3,2,5), y=(2,4,-5), and z=(1,6,7). Calculate:

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2. Sep 3, 2009

### NJunJie

I view such qns playing with 'arrays' and 'susbstituion'.

Generally, i will view it this way:-

x is an array of (-3,2,5)
y is an array of (2,4,-5)

first part is u do the summation first - i call it inner.

Inner: (-3)(2) + (2)(4)

Then you do the Products - i call it outer.

But is your question complete? Theres no 'j' in your formulaes pasted.

3. Sep 3, 2009

### HallsofIvy

Staff Emeritus
What you have written,
$$\prod_{j= 1}^3\sum_{i=1}^2 x_iy_i$$ and
$$\sum_{j=1}^3\prod_{i=1}^2 x_iy_i$$
are just
$$\prod_{j=1}^3(x_1y_1+ x_2y_2+ x_3y_3)= \prod_{j=1}^3((-3)(2)+ (2)(4)+ (5)(-5))= \prod_{j=1}^3(-6+ 8- 10)= 3(8)= 24$$
and
$$\sum{j= 1}^3((x_1y_1)(x_2y_2))= \sum_{j=1}^3 (-3)(2)(2)(4)= \sum_{j=1}^3 48= 3(48)= 144$$

But I suspect you meant
$$\prod_{j=1}^3\sum_{i= 1}^2 x_iy_j$$ and
$$\sum{j=1}^3\Pi_{i=1}^2 x_iy_j$$

The first of those is
$$\prod_{j=1}^3(x_1+ x_2)y_j= (x_1+ x_2)\prod_{j=1}^3y_i= (x_1+ x_2)(y_1y_2y_3)$$
surely you can do that arithmetic yourself.