Question

Use mathematical induction to prove that the statement is true for every positive integer n.

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

Answer 1

inductive and deductive reasoning.

Answer 2

let it is true for n=1
so
8(1)=4(1)(1+1)
8=8
fist condition is true
now let
its true for n=k
8 + 16 + 24 + . . . + 8k = 4k(k + 1)
we have to show that it is true for n=k+1

Answer 3

sorry page was reloaded :(
let me try again.

Answer 4

ok

Answer 5

now
when n=k+1
we need to show that
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
because
8 + 16 + 24 + . . . + 8k=4k(k+1)
simplify the 1
=4k(k + 1)+8(K+1)
=4k(k+1)+8k+8
=4k^2+4k+8k+8
=4k^2+12k+8
take 4 common

=4(k^2+3k+2)
=4(k+1)((k+2)
or
=4(k+1)((k+1)+1)
hence proved

Answer 6

very nice
you should figure out for yourself how this question arose

Answer 7

mathematical induction shows that the statement is true for every positive integer n.

Answer 8

\[8 + 16 + 24 + . . . + 8n = 4n(n + 1)
\]
\[8(1+2+3+...+n)=4n(n+1)\]
\[8\frac{n(n+1)}{2}=4n(n+1)\checkmark \]

Answer 9

so first u do n=1, then n=k+1. is that it?

Answer 10

it is a wee bit more complicated than that
first you check \(n=1\) yes
then you prove that IF it is true for \(n=k\) THEN it is true for \(n=k+1\)

Answer 11

if you are doing these for the first time you should write down precisely the IF part and the THEN part, so see what you know, and where you are headed

Answer 12

yes first show that it is true for n=1
then assume it is true for any naturals no k.
then show that it is true for n=k+1
bcz if it is true for n=k then should be true for the right next no .
if it is true for example for 1,2,3,4 then it should be true for 5 also.

Answer 13

in this example, the "if" part is
\[8 + 16 + 24 + . . . + 8k=4k(k+1)\]

Answer 14

this part you get to assume is true. that is called the "inductive hypothesis" that is is true for \(n=k\) and as @sami-21 wrote, you need to show that given the above, it follows that
\[8 + 16 + 24 + . . . + 8k +8(K+1)=4(k+1)((k+1)+1)\] here, he replaced \(n\) by \(k+1\)

Answer 15

at the risk of repeating myself, don't fly by the seat of your pants.
write down exactly what the inductive hypothesis is, so you have it in front of you
then write down what you are trying to prove

find the inductive part in what you are trying to prove, which is fairly easy in summation formulas, it is usually the sum up until the last term

replace that part by the formula you assume to be true
it is algebra from there on in

Answer 16

\[\overbrace{8+16+24+...+8k}+8(k+1)\] by induction you get to replace the first part by \(4k(k+1)\)

Answer 17

so after you have checked n=1 and you have proved n=k+1 that is it? or are they more steps?

Answer 18

it completes the proof.
steps are
1)show that it is true for n=1
2)assume it is true for n=k (hypothesis)
3)show that it is true for next no k+1
if it is true then it follows that it will be true for any k.

Answer 19

thnks!!!