Summation question

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Mckanic

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Looking for help modeling how to do a summation.

(y1+1) + (y1+y2+2) + (y1+y2+y3+3) > 1

The addition of the yi to each group is got me stumped.

I know the basic is

Sum i to 3 (yi + i) >1
 
Mckanic said:
Looking for help modeling how to do a summation.
\((y_1+1) + (y_1+y_2+2) + (y_1+y_2+y_3+3) > 1\)
Click to expand...
See how I eddied your notation.
That adds to \(3y_1+2y_2+y_3+6>1\)
 
Mckanic said:
Looking for help modeling how to do a summation.

(y1+1) + (y1+y2+2) + (y1+y2+y3+3) > 1

The addition of the yi to each group is got me stumped.

I know the basic is

Sum i to 3 (yi + i) >1
Click to expand...
Your goal is not clear. What do you mean by "how to do a summation"? Do you want to add these up, as pka did, or perhaps to write it using sigma notation, as you appear to be trying to do? Or something else?

You can represent the kth term on your left-hand side by [MATH]\left(\sum_{i=1}^k y_i\right) + k[/MATH], or [MATH]\sum_{i=1}^k(y_i + 1)[/MATH]. So the entire expression is [MATH]\sum_{k=1}^3\sum_{i=1}^k(y_i + 1)[/MATH].
 
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Dr.Peterson said:
Your goal is not clear. What do you mean by "how to do a summation"? Do you want to add these up, as pka did, or perhaps to write it using sigma notation, as you appear to be trying to do? Or something else?

You can represent the kth term on your left-hand side by [MATH]\left(\sum_{i=1}^k y_i\right) + k[/MATH], or [MATH]\sum_{i=1}^k(y_i + 1)[/MATH]. So the entire expression is [MATH]\sum_{k=1}^3\sum_{i=1}^k(y_i + 1)[/MATH].
Click to expand...
Your response is what I was looking for. Sorry for confusion.
 
Mckanic said:
Trying to figure out how this expands out to my original equation.
Click to expand...
\(\sum_{k=1}^3\sum_{i=1}^k(y_i + 1)\)
\(\begin{array}{*{20}{l}}
{k = 1}&{\sum\limits_{j = 1}^1 {({y_j} + 1) = } }&{({y_1} + 1)} \\
{k = 2}&{\sum\limits_{j = 1}^2 {({y_j} + 1) = } }&{({y_1} + 1) + ({y_2} + 1)} \\
{k = 3}&{\sum\limits_{j = 1}^3 {({y_j} + 1) = } }&{({y_1} + 1) + ({y_2} + 1) + ({y_3} + 1)}
\end{array}\)
 
Ok

I am expanding wrong I guess.

solving the second sum I thought would give me (y1+1) + (yk+1).

now solving the first sum {(y1+1)+(y1+1)} + {(y1+1)+(y2+1)} + {(y1+1)+(y3+1)}

(2y1+2) + (y1 +y2 +2) + (y1 + y3 + 2)

so that is 4y1 + y2 + y3 +6
 
pka

I understand that. Understand now what I was doing wrong. Thanks to all of you who replied.
 
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