1+6+11+...+(5p-4)
Last edited by Drongoski; 18 Jun 2017 at 9:50 AM.
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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?
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.
Last edited by Drongoski; 18 Jun 2017 at 1:52 PM.
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In simple terms, you need to recognise that the sequence is arithmetic, which can be proven by checking that T_{2} - T_{1} = T_{3} - T_{2}.
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.
Last edited by bujolover; 18 Jun 2017 at 1:07 PM.
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