Question

write the terms of the series and find their sum

Answer 1

hmmm...

Answer 2

@turtleluver

Answer 3

@agent0smith

Answer 4

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

Answer 5

@ehuman

Answer 6

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

plug in each value of k, one at a time, from k=1 to k=4

Like this, for k=1 and k=2...
(1+3)^2 = ...
(2+3)^2 = ...
do the same for k =3 and k=4, then add all the results together, that's your sum.

Answer 7

@agent0smith i came up with 1^2+3=4
4^2+3=19
=23

Answer 8

another way .. maybe more advanced:

\[\sum_{k=1}^{4} (k+3)^{2}\]
\[\sum_{k=1}^{4} (k^2+6k+9)\]
\[\sum_{k=1}^{4} k^2+6\sum_{k=1}^{4} k+9\sum_{k=1}^{4} \]
\[\sum_{k=1}^{4} k^2+6\frac{4(5)}{2}+9(4) \]
lol, i forget at the moment the k^2 setup

Answer 9

sum k^2 = 1+4+9+16
= 30

Answer 10

@Avon you need to follow order of operations... do everything in parentheses first

try it again.

(1+3)^2 = ...
(2+3)^2 = ...
(3+3)^2 = ...
(4+3)^2 = ...

Answer 11

16,25,36,49 @agent0smith

Answer 12

126 final answer

Answer 13

@agent0smith thanks for ya help

Answer 14

Looks right.