Use sigma notation to represent each of the following (1 Viewer)

poppyisloveinthewind · Jun 18, 2017

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

Drongoski · Jun 18, 2017

poppyisloveinthewind said:
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 · Jun 18, 2017

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

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 · Jun 18, 2017

poppyisloveinthewind said:
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 · Jun 18, 2017

Drongoski said:
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 · Jun 18, 2017

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 · Jun 18, 2017

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 · Jun 18, 2017

poppyisloveinthewind said:
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 · Jun 18, 2017

poppyisloveinthewind said:
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.

Hence, no trial and error is needed AT ALL.

Use sigma notation to represent each of the following (1 Viewer)

poppyisloveinthewind

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Drongoski

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poppyisloveinthewind

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Drongoski

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poppyisloveinthewind

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poppyisloveinthewind

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Drongoski

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InteGrand

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bujolover

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