Question

Write the terms of the series and find their sum

4E (k+8)^2
K=1

Answer 1

that E : \[\Sigma \]

Answer 2

lol

Answer 3

\[\sum_{k=1}^4(k+8)^2\]

Answer 4

YES that!

Answer 5

replace \(k\) by 1 then by 2 then by 3, then by 4
put plus signs between them

Answer 6

\[
\sum_{k=1}^4(k+8)^2
\]\(i=k+8\implies k=i-8\)\[
\sum_{i-8=1}^{i-8=4}(i)^2=\sum_{i=9}^{12}i^2
\]

Answer 7

\[\sum_{k=1}^4(k+8)^2=(1+8)^2+(2+8)^2+(3+8)^2+(4+8)^2\]

Answer 8

more easily written as
\[9^2+10^2+11^2+12^2\] then take out a calculator and add

Answer 9

so 64,100,121,144

Answer 10

\[
\sum_{i=9}^{12}i^2 = \sum_{i=0}^{12}i^2 - \sum_{i=0}^{8}i^2
\]

Answer 11

not 64, 81

Answer 12

i get 466
http://www.wolframalpha.com/input/?i=9^2%2B10^2%2B11^2%2B12^2

Answer 13

me too.

Answer 14

yay

Answer 15

so the terms are 81,100,121,144 and all that means that equation equals 446