Help! induction question (1 Viewer)

[maveric31]

really need help with this... before tomorrow:S

"use mathematical induction to show that:
1^3 + 2^3 +3^3... +n^3 = (1+2+3... +n)^2"

plz help

 

[shinn]

For n = 1

Thus, the result is true for n = 1.

Assume the result is true for n = k, i.e.
(I)

We need to show the result is true for n = k+1, i.e.
(II)

Now,

, by (I)
, sum of an AP series
, from factorising (k+1)^2


Also consider,

, sum of an AP series


Hence, LHS(II) = RHS(II).
So the result is true for n = k + 1 if it is also true for n = k.

Therefore, by mathematical induction, the result is true for all positive integers of n.

 

[shaon0]

NOTE: (1+...+k)=(k(k+1))/2 by sum of Arithmetic Progression.

First case for n=1 is trivial

Let n=k:
1^3+...+k^3=(1+...+k)^2

Let n=k+1:
LHS=1^3+...+k^3+(k+1)^3
=(1+...+k)^2+(k+1)^3
=k^2(k+1)^2/4+(k+1)^3
=RHS

RHS=((1+...+k)+k+1)^2
=(1+...+k)^2+2(k+1)(1+...+k)+(k+1)^2
=k^2(k+1)^2/4+(k+1)(k(k+1)+(k+1))
=k^2(k+1)^2/4+(k+1)^2.(k+1)
=k^2(k+1)^2/4+(k+1)^3

Above post is wrong as his case doesn't work for n=2 and has misinterpreted the question:
ie 1^3+2^3 =/= (1+2)^3

 

[Trebla]

NOTE: (1+...+k)=(k(k+1))/2 by sum of Arithmetic Progression.

First case for n=1 is trivial

Let n=k:
1^3+...+k^3=(1+...+k)^2

Let n=k+1:
LHS=1^3+...+k^3+(k+1)^3
=(1+...+k)^2+(k+1)^3
=k^2(k+1)^2/4+(k+1)^3
=RHS

RHS=((1+...+k)+k+1)^2
=(1+...+k)^2+2(k+1)(1+...+k)+(k+1)^2
=k^2(k+1)^2/4+(k+1)(k(k+1)+(k+1))
=k^2(k+1)^2/4+(k+1)^2.(k+1)
=k^2(k+1)^2/4+(k+1)^3

Above post is wrong as his case doesn't work for n=2 and has misinterpreted the question:
ie 1^3+2^3 =/= (1+2)^3

Should say ASSUME n = k is true.

 

[maveric31]

Thanks for the replies guys really appreciate it

 

[shaon0]

Should say ASSUME n = k is true.

Yeah, sorry. Usually i say: Assume < statement> (for n=k) but i didn't think it was necessary as i'm just writing out the proof for a person and not formally doing it in a test.

 

[shinn]

Yea thanks for pointing that out, i accidentally did a did a typo (1+...+k)^3, suppose to be (1+....+k)^2. fixed.