Question

Write the sum using summation notation, assuming the suggested pattern continues.

-4 + 5 + 14 + 23 + ... + 131

Answer 1

Write the sum using summation notation, assuming the suggested pattern continues.

16 + 25 + 36 + 49 + ... + n2 + ...

Answer 2

\[- 4 + 5 + 14 + 23 + ... + 131 = \sum_{n = 0}^{15} (-4) + n\cdot 9\]

Answer 3

\[16 + 25 + 36 + 49 + ... = \sum_{n=4}^{\infty}n²\]

Answer 4

Write the sum using summation notation, assuming the suggested pattern continues.

8 - 40 + 200 - 1000 + ...

Answer 5

\[8 - 40 + 200 - 1000 + ... = \sum_{n=0}^{\infty} (-1)^n \cdot 8 \cdot 5^n\]

Answer 6

maybe you prefer writing it as\[8 \cdot \sum_{n=0}^{\infty}(-5)^n\]

Answer 7

Find the sum of the first 12 terms of the sequence. Show all work for full credit.

1, -4, -9, -14, . . .

Answer 8

Obviously, our sequence can be written as\[a_n = 1 - n\cdot 5\] It follows that \[\sum_{n=0}^{11}a_n = \sum_{n=0}^{11}(1 - n\cdot 5) = \sum_{n=0}^{11} 1 - \sum_{n=0}^{11}n\cdot 5 = 11 - 5\cdot \sum_{n=0}^{11} n\] There is a more elegant way to do this, but I you're probably better off if you just calculate the following sum "manually", by just adding up all the terms up to eleven.\[\sum_{n=0}^{11}n = 1 + 2 + ... + 11 = 66\]Hence the sum is \[\sum_{n=0}^{11}a_n = 11 - 5\cdot \sum_{n=0}^{11}n = 11 - 5\cdot 66 = - 319 \]

Answer 9

Thank you so much for all of your help! :)

Answer 10

we can all see a difference of 9.
and that the first term (to which you are adding 9) is -4.

131 is apparently the end of the pattern, so we need to determine whichTH is 131.
\(\large\color{black}{131=-4+9(n-1)}\)
\(\large\color{black}{135=9(n-1)}\)
\(\large\color{black}{15=n-1}\)
\(\large\color{black}{16=n}\)

so \(\large\color{black}{131}\) is the \(\large\color{black}{16}\)th term.

Answer 11

So my final answer would be: \(\large\color{blueviolet}{ \displaystyle \sum_{ n=0 }^{ 15 } ~ (9n-4)}\)

Answer 12

And no, I don't think it is for an infinite number of terms.