Finding the sum of 1^3 + 2^3 + + n^3 by induction

[jonroberts74]

1^3+2^3+...+n^3 = \left[ \frac{n(n+1)}{2}\right]^2; n\ge 1

P(1) = 1^3 = \frac{8}{8} = 1

P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2 (induction hypothesis)

P(k+1) = 1^3+...+k^3+(k+1)^3 = \left[\frac{(k+1)(k+2)}{2}\right]^2

I start getting stuck here
I foiled it out then let m = P(k)

\left[ m + \frac{2(k+1)}{2}\right]^2

 

[Quesadilla]

You probably meant $m = \sqrt{P(k)}$. Now expand the square and use the induction hypothesis.

 

[ehild]

You assume that P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2 To get P(k+1), you have to add (k+1)^3 to P(k). P(k+1)=P(k)+(k+1)^3=\left[ \frac{k(k+1)}{2}\right]^2+(k+1)^3 Write it out, and show that it is equal to P(k+1) = \left[ \frac{(k+1)(k+2)}{2}\right]^2

Spoiler

pull out (k+1)²

ehild

 

[jonroberts74]

You assume that P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2 To get P(k+1), you have to add (k+1)^3 to P(k). P(k+1)=P(k)+(k+1)^3=\left[ \frac{k(k+1)}{2}\right]^2+(k+1)^3 Write it out, and show that it is equal to P(k+1) = \left[ \frac{(k+1)(k+2)}{2}\right]^2

Spoiler

pull out (k+1)²

ehild

\left[\frac{(k+1)(k+2)}{2}\right]^2 = \left[ \frac{k(k+1)}{2}\right]^2+(k+1)^3

(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{k^2}{2^2} + (k+1)\right]

(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{k^2+4k+4}{2^2}\right]

(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{(k+2)^2}{2^2}\right]

?

 

[ehild]

\left[\frac{(k+1)(k+2)}{2}\right]^2 = \left[ \frac{k(k+1)}{2}\right]^2+(k+1)^3

(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{k^2}{2^2} + (k+1)\right]

(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{k^2+4k+4}{2^2}\right]

(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{(k+2)^2}{2^2}\right]

?

Apply that a²b²=(ab)²

(k+1)^2 \left[\frac{(k+2)}{2}\right]^2=\left[(k+1)\frac{(k+2)}{2}\right]^2= P(k+1)

is it familiar? Is it what you wanted to arrive at?


ehild