Question

Discrete math:

Use induction to show that for all positive integers n
(a) 13 + 23 + 33 + . . . + n
3 = (n(n + 1)/2)2
.
(b) 1 · 1! + 2 · 2! + . . . + n · n! = (n + 1)! − 1
(c) if n > 6, then 3n < n!

I did some work for a, and b but im stuck:

a--> assume for 1
1^3 = (1(1+1)/2)^2 which is correct
Hypothesis: 1^3+2^3+...+k^3= (k(k+1)/2)^2

Induction step: 1^3+2^3+...+k^3+(k^3+1)= (k(k+1)/2)^2 + (k^3 + 1)

I started with the RHS and got this:
(k^4 + 6k^3 + k^2)/4 + 1 but can't seem to get back to the original rule
for b:
both Hypothesis and n(1) work

Answer 1

works, so the Induction step left me with 2(k+1)! +k which I can't seem to bring back to the original rule too, any help? :/