Question

Write the sum using summation notation, assuming the suggested pattern continues. -8 - 3 + 2 + 7 + ... + 67

summation of the quantity negative eight plus five n from n equals zero to fifteen

summation of negative forty times n from n equals zero to infinity

summation of negative forty times n from n equals zero to fifteen

summation of the quantity negative eight plus five n from n equals zero to infinity

Answer 1

-8 - 3 + 2 + 7 + ... + 67 hmm... so the first term's value is -8. If you notice, it went from -8, to -3 and then +2 and so on.... you can always get the "common difference" d by simply getting the difference of two terms, so d = +2 -(- 3) ----> d = +2+3 ----> d = 5.

now.. the last value is 67.. but what term is that anyway?

well, let's use the arithmetic sequence equation to see who that is.


`\bf n^(th)\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ a_1=-8\\ d=5\\ a_n=67 \end{cases} \\\\\\ a_n=-8+(n-1)5\implies 67=-8+(n-1)5 \\\\\\ 67=-8+5n-5\implies 67=-13+5n\implies 80=5n \\\\\\ \cfrac{80}{5}=n\implies \boxed{16=n}`

so, is the 16th term alrite.. .hmmm, so it begins with -8, then keeps on adding 5 16 times


`\bf \sum\limits_(n=1)^(16)~-8+(n-1)5\implies \sum\limits_(n=1)^(16)~5n-13`

both forms are the same, so.. hmm it depends on how simplified you need it.