Thread: Proof by Induction View Single Post
 Emeritus Sci Advisor PF Gold P: 5,532 OK, here is the solution in all the gory details. P(k)=Σj=1k=(1/4)k2(k+1)2 Expanding the rightmost side, I get: P(k)=Σj=13=(1/4)(k4+2k3+k2) Now, look at P(k+1): Σj=1k+1j3=(1/4)(k+1)2(k+2)2 I am going to manipulate both sides of this statement, one at a time: Isolate (k+1)st Term On Left Side: Σj=1k+1=Σjkj3+(k+1)3 Expand Right Side: (1/4)(k4+6k3+13k2+12k+4) So, I have: Σj=1kj3+(k+1)3=(1/4)(k4+6k3+13k2+12k+4) Now, remember P(k)? Assume it is true, and subtract it from P(k+1). It will be handy if you subtract the left side of P(k) from the left side of P(k+1), and do likewise with their right sides. I am going to denote both sides of P(k) with the color blue. Then you get: Σj=1kj3+(k+1)3-Σj=1kj3=(1/4)(k4+6k3+13k2+12k+4)-(1/4)(k4+2k3+k2) And finally, (k+1)3=k3+3k2+3k+1 which is an identity. Therefore, P(k)-->P(k+1). edit: fixed superscript bracket