Question

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

-10 - 2 + 6 + 14 + ... + 110

Answer 1

l=a+(n-1)d
a=-10
l=110
d=-2-(-10)=8
110=-10+(n-1)8=-10+8n-8=8n-18
110+10=(n-1)8
120/8=n-1
n-1=15
n=15+1=16
solve
\[\sum_{1}^{16}\left( 8n-18 \right)\]

Answer 2

a1 = -10
a2 = -10 + 8 = a1 + 8
a3 = -10 + 8 + 8 = a1 + 8(2 - 1)
a4 = -10 + 8 + 8 + 8 = a1 + 8(3 - 1)
an = a1 + 8(n - 1)
an = -10 + 8(n - 1) = -10 + 8n - 8
an = -18 + 8n
an = 8n - 18

\(\Large \sum \limits_{n = 11}^{16} ( 8n-18 )\)

Answer 3

These are my answer choices:
summation of negative eighty times n from n equals zero to fifteen
summation of the quantity negative ten plus eight n from n equals zero to fifteen
summation of the quantity negative ten plus eight n from n equals zero to infinity
summation of negative eighty times n from n equals zero to infinity

Answer 4

@xapproachesinfinity