Question

Find the sum of the geometric sequence.

1 divided by 3, 2 divided by 3, 4 divided by 3, 8 divided by 3, 16 divided by 3

31 divided by 3

10

2

31 divided by 15

Answer 1

@hoblos @eliassaab @ganeshie8

Answer 2

will give best answer & medal!

Answer 3

Here's the geometric sequence: 1/3 , 2/3 , 4/3 , 8/3, 16/3

Answer 4

\(\large \frac{1}{3}, \frac{2}{3}, \frac{4}{3}, \frac{8}{3}, \frac{16}{3}\)

Answer 5

the answer choices: A) 31/3
B) 10
C) 2
D) 31/15

Answer 6

either add them up manually

or

use partial series formula

Answer 7

kk! when I add them up manually i got 10 (:

Answer 8

is that the correct answer?

Answer 9

\(\large \frac{1}{3} + \frac{2}{3} + \frac{4}{3} + \frac{8}{3} + \frac{16}{3}\)


\(\large \frac{1}{3}\left(1 + 2 + 4 + 8 + 16\right)\)

Answer 10

try again

Answer 11

I still got 10
lol :(

Answer 12

\(\large \frac{1}{3} + \frac{2}{3} + \frac{4}{3} + \frac{8}{3} + \frac{16}{3}\)


\(\large \frac{1}{3}\left(1 + 2 + 4 + 8 + 16\right)\)

\(\large \frac{1}{3}\left(31\right)\)

Answer 13

Yes that's equal to 10.323

Answer 14

10.323 is NOT equal to 10

Answer 15

look at ur options,
u have 31/3 also right ?

Answer 16

Sorry my teacher makes us round our answers!

Answer 17

Ohhhhhhhhhhhhh good catch!

Answer 18

thank u SOO much!

Answer 19

:)

Answer 20

np :)

Answer 21

Use mathematical induction to prove the statement is true for all positive integers n.

8 + 16 + 24 + ... + 8n = 4n(n + 1)

Answer 22

@ganeshie8 can u help me with this one?

Answer 23

since u seem to definatley know what ur doing! lol

Answer 24

:)

Answer 25

3 steps :

1) show that the formula works for \(n = 1\)
2) assume the formula works for some \(n=k\)
3) show that it works for \(n=k+1\)

Answer 26

^^thats the basic setup for any induction proof

Answer 27

Can u show me the steps written out? I've never done this before

Answer 28

\(8 + 16 + 24 + ... + 8n = 4n(n + 1)\)

step 1 :
lets see if the formula works for \(n=1\) case

Answer 29

plugin \(n=1\), on right hand side formula
wat do u get ?

Answer 30

4(1)(2)
8 :)

Answer 31

so, n =1 case lies in our solution set for formula.

move to step 2

Answer 32

ok :)

Answer 33

that's the trickiest part

Answer 34

\(8 + 16 + 24 + ... + 8n = 4n(n + 1)\)

step 1 :
lets see if the formula works for \(n=1\) case
\(4*1(1+1) = 4(2) = 8\)

So, the formula is \(true\) for sum of \(1\) terms.

step 2 :
assume that the formula for works for \(n=k\)
then : \(8 + 16 + 24 + ... + 8k = 4k(k + 1)\)

Answer 35

actually, step2 is the easiest step in any induction proof :-
u just need to assume that the formula works for n = k, and replace n wid k in the formula... thats all.

move to step 3

Answer 36

oh ok!

Answer 37

step 3 can be a bit tricky,
as u need to show that the formula works for n = k+1, whenever it worked for n = k

Answer 38

its not that long process,
its just 3 steps :)

Answer 39

once u do 1-2 problems, u will start to like these proofs

Answer 40

got it :)

Answer 41

\(8 + 16 + 24 + ... + 8n = 4n(n + 1)\)

step 1 :
lets see if the formula works for \(n=1\) case
\(4*1(1+1) = 4(2) = 8\)

So, the formula is \(true\) for sum of \(1\) terms.

step 2 :
assume that the formula for works for \(n=k\)
then : \(8 + 16 + 24 + ... + 8k = 4k(k + 1)\)

step 3 :
sum of \(k+1\) terms is
\(8 + 16 + 24 + ... + 8(k+1)\)

Answer 42

you need to somehow show that it equals : \(4(\color{red}{k+1})(\color{red}{k+1}+1)\)

Answer 43

any idea, why we need to show that ?

Answer 44

hmm, its to prove the formula true for all positive integeres

Answer 45

I'm not sure

Answer 46

you wanted to prove : \(8 + 16 + 24 + ... + 8n = 4n(n + 1)\)

Answer 47

plugin n = (k+1)
wat do u get ?

Answer 48

The answer u posted in red above!

Answer 49

where ever u see "n", just blindly put "k+1"

Answer 50

4(k+1)(k+1+1)
:)

Answer 51

will do!

Answer 52

yup!

Answer 53

so,

\(8 + 16 + 24 + ... + 8(k+1) = 4(k+1)((k+1) + 1) \)

is the sum formula for k+1 terms

Answer 54

lets go back to the proof

Answer 55

ok :)

Answer 56

btw your an awesome teacher :)

Answer 57

\(8 + 16 + 24 + ... + 8n = 4n(n + 1)\)

step 1 :
lets see if the formula works for \(n=1\) case
\(4*1(1+1) = 4(2) = 8\)

So, the formula is \(true\) for sum of \(1\) terms.

step 2 :
assume that the formula for works for \(n=k\)
then : \(8 + 16 + 24 + ... + 8k = 4k(k + 1)\)

step 3 :
sum of \(k+1\) terms is
\(8+16+24+...+8(k+1) = 8+16+24+...+8k+8 \)

from step2 of induction hypothesis, this equals :
\( 4k(k+1)+8 \)

Answer 58

we're almost done,
just make sure you're at peace wid above^

Answer 59

next u need to show
\(4k(k+1) + 8\)

is same as

\(4(k+1)(k+1 + 1)\)

Answer 60

yes definatley at pace :) sorry if im slow im taking notes but i only have 10 mins left on the computer

Answer 61

\(4k(k+1) + 8\)

\(4k^2+4k + 8\)

\(4(k^2+k + 2)\)

Answer 62

factor the thing inside parenthesis,
wat do u get ?

Answer 63

is it factorable?

Answer 64

Oh you mean factor it out
4k^2 + 4k +8

Answer 65

yup,

factor below
\(k^2 + k + 2\)

Answer 66

Idk how to do that sorry :(

Answer 67

You can do it really quick and I will search how to do it later due to the low time amount

Answer 68

actually i made a mistake earlier

Answer 69

in step 3

Answer 70

whats the mistake?

Answer 71

\(8 + 16 + 24 + ... + 8n = 4n(n + 1)\)

step 1 :
lets see if the formula works for \(n=1\) case
\(4*1(1+1) = 4(2) = 8\)

So, the formula is \(true\) for sum of \(1\) terms.

step 2 :
assume that the formula for works for \(n=k\)
then : \(8 + 16 + 24 + ... + 8k = 4k(k + 1)\)

step 3 :
sum of \(k+1\) terms is
\(8+16+24+...8k+8(k+1) \)

from step2 of induction hypothesis, this equals :
\( 8+16+24+...8k+8(k+1) = 4k(k+1)+8(k+1) \)

\(\qquad ~~~~ = 4k^2 + 4k + 8k + 8\)

\(\qquad ~~~~ = 4k^2 + 12k + 8\)

\(\qquad ~~~~ = 4(k^2 + 3k + 2)\)

\(\qquad ~~~~ = 4(k+1)(k+2)\)

\(\qquad ~~~~ = 4(k+1)(k+1 + 1)\)

So, the formula works for \(n = k+1\) also.

Hence the formula works for all \(n \ge 1\)

Answer 72

thank u SOOO much! :)

Answer 73

^^complete proof above,
let me knw if smthng doesnt make sense..

Answer 74

will do!
it should since ur a professional at this :)

Answer 75

lol no,
u wil become professional after doing 2-3 proofs, trust me u wil get addicted to induction proofs once u feel how powerful they're lol

Answer 76

lol ok :) thank u so much for ur help, I wish i could give you best response more than once

Answer 77

:) you're wlcme !