Can Properties of Sigma Notation Prove Equivalence of Two Sums?

[songoku]

Homework Statement

Prove that:

\sum_{n=1}^{14} 10n = \sum_{n=1}^{7} (20n+70)

Homework Equations

properties of sigma notation

The Attempt at a Solution

I know several properties of sigma notation but none that I know can be used to prove this. I don't know how to change 10 n to 20 n

Thanks

 

[micromass]

What properties do you know of sigma notation?

 

[tiny-tim]

hint: there's exactly twice as many terms in one sum as in the other sum

 

[Ray Vickson]

Homework Statement

Prove that:

\sum_{n=1}^{14} 10n = \sum_{n=1}^{7} (20n+70)

Homework Equations

properties of sigma notation

The Attempt at a Solution

I know several properties of sigma notation but none that I know can be used to prove this. I don't know how to change 10 n to 20 n

Thanks

Sigma notation is a shorthand that helps you to write things in a more compact form, and is very useful once you have grasped what is being asked. However, if you have not yet understood what is being asked, or what has been written, the sigma notation is just getting in the way. It would be better in this case to write out in detail both sides of what was written above, but NOT using sigma notation. That way, you can understand what, exactly, you are being asked to do. After you understand these matters better, THEN you can start to switch back to using sigma notation, to save space, etc.

RGV

 

[songoku]

What properties do you know of sigma notation?

\sum_{n=1}^{r}c = cr ; c = constant
\sum_{n=1}^{r}c.p_n=c\sum_{n=1}^{r}p_n
\sum_{n=1}^{r} (p_n+q_n)=\sum_{n=1}^{r}p_n+\sum_{n=1}^{r}q_n
\sum_{n=1}^{r}p(n)=\sum_{n=1+s}^{r+s}p(n-s)
\sum_{n=1}^{r}p_n=\sum_{n=1}^{m}p_n+\sum_{n=m+1}^{r}p_n

hint: there's exactly twice as many terms in one sum as in the other sum

Hm...sorry I don't know how to use your hint...:shy:

Sigma notation is a shorthand that helps you to write things in a more compact form, and is very useful once you have grasped what is being asked. However, if you have not yet understood what is being asked, or what has been written, the sigma notation is just getting in the way. It would be better in this case to write out in detail both sides of what was written above, but NOT using sigma notation. That way, you can understand what, exactly, you are being asked to do. After you understand these matters better, THEN you can start to switch back to using sigma notation, to save space, etc.

RGV

Not sure what you mean but let me try

Expand the LHS:
10 + 20 + ... + 140
= (20 + 40 + 60 + 80 + ... + 140) + (10 + 30 + 50 + ... + 130)
=\sum_{n=1}^{7}20n + 70.7
=\sum_{n=1}^{7}(20n+70)

But I seriously doubt the validity of my work...

Thanks

 

[tiny-tim]

hi songoku!

Hm...sorry I don't know how to use your hint...:shy:

there's 14 terms on the left, and 7 on the right …

so try putting the 14 terms into 7 pairs of 2 terms, so that each pair adds to the correct amount

(eg, {1,2},{3,4}etc or {1,14}{2,13}etc or … )

 

[songoku]

hi songoku!


there's 14 terms on the left, and 7 on the right …

so try putting the 14 terms into 7 pairs of 2 terms, so that each pair adds to the correct amount

(eg, {1,2},{3,4}etc or {1,14}{2,13}etc or … )

hi tiny tim

So it is basically the same as what I have done in post #5?

Is there another way to prove it without expanding the sigma notation? I mean just using the properties such as changing the upper and lower bound or other properties

Thanks

 

[tiny-tim]

hi songoku!

So it is basically the same as what I have done in post #5?

you mean …

10 + 20 + ... + 140
= (20 + 40 + 60 + 80 + ... + 140) + (10 + 30 + 50 + ... + 130)
=\sum_{n=1}^{7}20n + 70.7

i don't see how you got that last line

try writing the LHS on one line (in numbers, in full), and the RHS on the next line, and then just connecting up anything that adds up

Is there another way to prove it without expanding the sigma notation? I mean just using the properties such as changing the upper and lower bound or other properties

yes, but you need to find what to do first before you can put in into proper maths

 

[songoku]

hi tiny-tim

hi songoku! you mean …i don't see how you got that last line

try writing the LHS on one line (in numbers, in full), and the RHS on the next line, and then just connecting up anything that adds up

\sum_{n=1}^{14}10n
= (20 + 40 + 60 + 80 + ... + 140) + (10 + 30 + 50 + ... + 130)
=\sum_{n=1}^{7}(20n) + (10+130) + (30+110)+(50+90)+70
=\sum_{n=1}^{7}(20n) + 140 + 140 + 140 + 70
=\sum_{n=1}^{7}(20) + 70 + 70 + 70 + 70 + 70 + 70 + 70
=\sum_{n=1}^{7}(20n) + 7.70
=\sum_{n=1}^{7}(20n+70)

Is this valid? And how to prove it without expanding?

Thanks

 

[tiny-tim]

hi songoku!

yes, that's certainly valid

to put that into symbolic language, try replacing the index n (from 1 to 14) by m (from 1 to 7), so that you use 2m for the even numbers and 2m±1 for the odd numbers

(another way to see the result, which i was thinking of at the start, is to match:
10 + 20 + 30 + 40 + 50 + 60 + 70 +
80 + 90 + 100+110+120+130+140

90 +110 + 130+150+170+190+210 )​

 

[songoku]

Hi tiny-tim

Sorry for late reply. Thanks a lot for all the help