Question

PLEASE HELP ME! Find the geometric partial sum of...

Answer 1

\[\sum_{j=1}^53(0.1)^{j-1}\] like that ?

Answer 2

yes

Answer 3

would you just plug in all values from 1-5 in the 3(0.1)^i-1

Answer 4

like 3(0.1)^1-1

Answer 5

pull the 3 out first and write
\[3\sum_{j=1}^5(0.1)^{j-1}\]

Answer 6

wait why?

Answer 7

you can multiply by 3 at the end after computing
\[\sum_{j=1}^{5}(0.1)^{j-1}\] which is actually much easier than it looks

Answer 8

oh why can you pull out the 3 outside the summation? that is the distributive law
multiplication distributes over addition
\[3a_1+3a_2+3a_3+3a_4+3a_5=3(a_1+a_2+a_3+a_4+a_5)\]

Answer 9

so how would you solve that. Im confused

Answer 10

so you might as well just add up
\[(0.1)^0+(0.1)^1+(0.1)^2+(0.1)^3+(0.1)^4+(0.1)^5\] and multiply the result by 3

Answer 11

like i said, this is much easier than it looks
first term is \((0.1)^0=1\) since anything to the power of zero is one
second term is \((0.1)^1=0.1\)
third term is \((0.1)^2=0.01\)
fourth term is \((0.1)^3=0.001\)
fifth term is ...

Answer 12

0.0001

Answer 13

when you add them you get
\(1.1111\)

Answer 14

ohh okayy! then multiply it by 3?

Answer 15

so the answer is 3.3333?

Answer 16

easy right? they are decimals that are all ones in different places
yes, then multiply by 3

Answer 17

yup

Answer 18

thankyou!

Answer 19

yw