Question

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

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

Answer 1

so you want to show \[\sum_{i=1}^{n}8i=4n(n+1) \]
let's see if the base case holds

Answer 2

i will let you test base case

Answer 3

The hardest part
is suppose to \[\sum_{i=1}^{k}8i=4k(k+1) \text{ for some integer } k>0\]
and then show
\[\sum_{i=1}^{k+1}8i=4(k+1)(k+2) \text{ for some integer } k>0\]

Answer 4

@freckles

Answer 5

looking over

Answer 6

8 + 16 + 24 + . . . + 8k +8(K+1)=4(k+1)((k+1)+1)
8 + 16 + 24 + . . . + 8k +8(K+1)= 4k(k + 1)+8(K+1).....1
what happened between these two lines

Answer 7

or what happened between
8 + 16 + 24 + . . . + 8k = 4k(k + 1)
8 + 16 + 24 + . . . + 8k +8(K+1)=4(k+1)((k+1)+1)

Answer 8

it looks like you added 8(k+1) on one side did you do that to the other side?

Answer 9

\[8+16+24+ \cdots +8k+8(k+1)=4k(k+1)+8(k+1)?\]

Answer 10

8 + 16 + 24 + . . . + 8k +8(K+1)=4(k+1)((k+1)+1)
I think I will ignore this line
I think ...1 you are just calling that equation 1 in
8 + 16 + 24 + . . . + 8k +8(K+1)= 4k(k + 1)+8(K+1).....1
so that is a good start for this

Answer 11

and it looks like you took it the long way around

Answer 12

this is setup already for factoring by grouping

Answer 13

4k(k+1)+8(k+1)
factor the common factor 4(k+1) out

Answer 14

and I get what you did
you were just starting what you wanted to show before you started showing it
i got confused earlier
your proof looks fine

Answer 15

I'm just saying you could reduce the steps by factoring by grouping

Answer 16

Ah, yes thats true. I understand, thank you

Answer 17

np