mathematical induction question: 1x1! + 2x2! + ... + nxn! = (n+1)!-1

Question

I know the solution to this question:

test basis step:
p(1)= 1x1!=(1+1)!-1
1x1 = (2)!-1
1 = 2-1, true

assume n=k
p(k) = 1x1! + 2x2! + ... + kxk! = (k+1)!-1

proof that if p(k) then p(k+1)
p(k+1) = 1x1! + 2x2! + ... + kxk! + (k+1)x(K+1)! = [(k+1)+1]!-1

here is where I get lost how did we get from:
p(k+1) = (k+1)!-1 + (k+1)x(k+1)! = (k+2)!-1
to
p(k+1) = (k+1)! x (1+k+1)-1 = (k+2)!-1

As to your specific question... "how did we get from..." Recall that $\color{red}{b} + a\color{red}{b} = 1\cdot \color{red}{b} + a\cdot \color{red}{b} = (1+a)\cdot \color{red}{b}$. Also, recall that $(a+1)\cdot a! = (a+1)!$. Here... $\color{red}{(k+1)!}+(k+1)\color{red}{(k+1)!} = (1+(k+1))\color{red}{(k+1)!} = (k+2)\cdot (k+1)! = (k+2)!$ — JMoravitz, Jul 01 '21 at 13:31
Also, highly recommended reading: How to write a clear induction proof? You should not be writing what you hope to be true and through manipulation attempt to arrive at a tautology... you should be starting with only one side and through a chain of equalities reach the desired second side. — JMoravitz, Jul 01 '21 at 13:33
Dear JMoravitz, thank you very much for the clear expalnation, kind regards Edward — Edward Kenway, Jul 02 '21 at 09:37

score 1 · Accepted Answer · answered Jul 01 '21 at 13:34

1

$$p(k+1)=(k+1)!-1 + (k+1)\cdot (k+1)!$$ $$=(k+1)!\left[1+k+1\right] - 1$$ $$=(k+1)!\left(k+2\right) - 1$$ $$=(1\cdot2\cdot3.....(k+1)(k+2))-1$$ $$=(k+2)!-1$$

answered Jul 01 '21 at 13:34

Smriti Sivakumar

999

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Do not answer questions that are duplicate. If you have a proof that is different from those givin in the earlier question, post it here. – Gary Jul 01 '21 at 13:38
Ah, answered it before the duplicate tag was placed, my bad!! – Smriti Sivakumar Jul 02 '21 at 03:35
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Dear Smitri Sivakumar, thank you for the clear and detailed answer Edward – Edward Kenway Jul 02 '21 at 09:43

mathematical induction question: 1x1! + 2x2! + ... + nxn! = (n+1)!-1

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