Question

Find (see drawing)
a 2,054
b 2,142
c 2,212
d 2,119

Answer 1

\[\sum_{i=9}^{22} 10i + 3\]

Answer 2

Use the properties of summations.
Namely,
\[\sum_{i}ai=a\sum_{i}\]

Answer 3

I don't know how to do that

Answer 4

Actually, this needs to be clarified. Is the sum:
\[
\sum_{i=9}^{22}\left(10i+3\right)
\]
or
\[
\left(\sum_{i=9}^{22}10i \right)+3
\]

Answer 5

The first one

Answer 6

Ok. Then here's what you do. Use these two facts:
\[
\begin{align}
\sum_{i}ca_{i}&=c\sum_{i}a_{i}\\
\sum_{i}\left(a_{i}+b_{i}\right)&=\sum_{i}a_{i}+\sum_{i}b_{i}
\end{align}
\]

The first means that you can pull out constant factors. The second means that summation is a linear operator. So, via the second rule:

\[
\sum_{9 \leq i \leq 22}\left(10i+3\right)=\sum_{9 \leq i \leq 22}10i+\sum_{9 \leq i \leq 22}3
\]

Via the first rule,

\[\sum_{9 \leq i \leq 22}10i+\sum_{9 \leq i \leq 22}3=10\sum_{9 \leq i \leq 22}i+3\sum_{9 \leq i \leq 22}1\]

You also have to change the range of summation to make things easier. Use \(1 \leq i \leq 14\) rather than \(9 \leq i \leq 22\). (They are both equivalent.)

So:

\[10\sum_{9 \leq i \leq 22}i+3\sum_{9 \leq i \leq 22}1=10\sum_{1 \leq i \leq 14}i+3\sum_{1 \leq i \leq 14}1\]

Can you take it from here? The majority of the work is done.

Answer 7

Wow!! I am still kinda lot but that was a lot so thanks!!

Answer 8

lost*

Answer 9

You're welcome. Summation is not easy to deal with. You would have to do it manually if you didn't know algebraic tricks. Doing it manually is tedious, even if it is easy for some sums.

Answer 10

so, after I have it like you said, then what do I do?

Answer 11

It kinda becomes trivial here. You have two methods:
Sum the sum by hand: \(1+2+3+4+\dots+14\). Or, use the formula for that particular sum. For the second sum, it is just \(1\) added \(14\) times. So it is trivially \(14\). From there, it's just a matter of multiplication and addition. :D

Answer 12

Wait, so do I use the first one?

Answer 13

You have to choose one. It's up to you. If you have the formula for
\[\sum_{1 \leq i \leq n}i\]
then you can use that (I prefer that). If you do not, you have to do it by hand. (That's what the first part meant).

The other sum,
\[\sum_{1 \leq i \leq 14}1\]
is just \(14\).

Answer 14

Will they both give me the same answer?

Answer 15

Yup. :) That's why you can do either one.

Answer 16

Thanks so, after i added 1+2+3+4+5+6+7+8+9+10+11+12+13+14 I got 105, then what

Answer 17

Put it back into the last equation. Then put in \(14\) where the other sum was. Then solve.

Answer 18

Which other sum?

Answer 19

The first one is:
\[
\sum_{1 \leq i \leq 14}i=105\]
The second is
\[\sum_{1 \leq i \leq 14}1=14\]

Answer 20

Okay so If I use the first one , wait why am I using 14 not 22?

Answer 21

I got it, the answer should be 2212

Answer 22

Yup! Great job.

Answer 23

I apologize, though. I made one mistake:
When adjusting the bounds, you have to change the first sum:
\[\sum_{9 \leq i \leq 22}i=\sum_{1 \leq i \leq 14}(i+8)=\sum_{1 \leq i \leq 14}i+8\sum_{1 \leq i \leq 14}1=105+8\cdot 14=105+112=217\]

Sorry.

Answer 24

No worries! I worked through it~ Thanks so much!!

Answer 25

You're welcome! I'm glad I could help. You may find the Wikipedia article helpful on future problems: http://en.wikipedia.org/wiki/Summation#Identities That'll give you all kinds of cool manipulations. :)

Answer 26

Thanks!!