Use sigma notation to represent each of the following

**poppyisloveinthewind** · 18 Jun 2017, 7:05 AM

1+6+11+...+(5p-4)

**Drongoski** · 18 Jun 2017, 9:44 AM

Originally Posted by poppyisloveinthewind

1+6+11+...+(5p-4)

$\sum _{i=1} ^p 5i - 4$

or, if you prefer:

$\sum _{k=1} ^p 5k - 4$

or even:

$\sum _{j= -2} ^{p-3} 5j+11$

**poppyisloveinthewind** · 18 Jun 2017, 9:53 AM

Originally Posted by Drongoski

$\sum _{i=1} ^p 5i - 4$

or, if you prefer:

$\sum _{k=1} ^p 5k - 4$

or even:

$\sum _{j= -2} ^{p-3} 5j+11$

was this done through trial and error

**Drongoski** · 18 Jun 2017, 10:31 AM

Originally Posted by poppyisloveinthewind

was this done through trial and error

No no - not by trial-&-error. As I am very well-versed with the use of the sigma notation, this is nothing to me. It may take you a while to get used to it. Don't be discouraged.

**poppyisloveinthewind** · 18 Jun 2017, 10:36 AM

Originally Posted by Drongoski

No no - not by trial-&-error. As I am very well-versed with the use of the sigma notation, this is nothing to me. It may take you a while to get used to it. Don't be discouraged.

I thought it was +5 for some reason and hence couldnt solve it

**poppyisloveinthewind** · 18 Jun 2017, 10:39 AM

right i think i get it now, the answer is in the question and by using the last value you have to do trial and error, am i right?

**Drongoski** · 18 Jun 2017, 10:57 AM

What trial & error?

My 1st and 2nd versions are simply a shorthand for:

(5x1 - 4) + (5x2 - 4) + (5x3 - 4) + . . . + (5xp - 4)

In this case we can express any term of the sum by a formula based on the term number. For this question, the general term is 5n-4 for the nth term. On the other hand, if you want to express 1 + 67 -5 + 1 + 87 +9 - 92 + 109 -55 -13 + 27 + 16 you won't be able to write this down in a sigma notation.

**InteGrand** · 18 Jun 2017, 11:40 AM

Originally Posted by poppyisloveinthewind

was this done through trial and error

$\noindent The fact that the terms form an arithmetic progression with \underline{common difference $5$} tells us that the general term is of the form $5k+b$ for some constant $b$ (where $k$ will be our summation index). Then depending on what you want to start your $k$ as, you can find $b$. E.g. if you want to start with $k = 0$, then the $k = 0$ term (starting term in the sum) being $1$ means that $b$ will be $1$ (since we need $5k + b = 1$ when $k = 0$). This makes the general term $5k+1$ (as $b = 1$). Thus the final term will be when $5k+1 = 5p - 4$, i.e. $k = p - 1$.$

**bujolover** · 18 Jun 2017, 1:04 PM

Originally Posted by poppyisloveinthewind

right i think i get it now, the answer is in the question and by using the last value you have to do trial and error, am i right?

In simple terms, you need to recognise that the sequence is arithmetic, which can be proven by checking that T₂ - T₁ = T₃ - T₂.
Ok, now that is done, we need to find T_n. Using the formula for an arithmetic sequence, T_n = a + d(n - 1), we get T_n = 1 + 5(n - 1) = 5n - 4.

Now, we just need to fill in the sigma part! So, on the bottom, we get n = 1, because 1 is the first term (sub n = 1 into T_n to see why). On the top, since it is clear that the last term of the sequence that's being added is the p^th term, p is placed on the top of the sigma. Finally, we just put 5n - 4 right next to the sigma (you can put brackets around 5n - 4 if you want to make it unambiguous).
See below for the answer.

Sigma Question.gif

Hence, no trial and error is needed AT ALL.