(New page: I don't even know how to solve the first problem. I tried it this way. Base step: P(0) 0*0 = 0 (0+1)! -1 =0 It is true Inductive step: Assume P(k) is true then P(k+1) is also true. ...) |
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From the first equation | From the first equation | ||
| − | (1*1! + 2+2! +…………………………..+ k*k!) +(k+1)*(k+1)! = (k+1)! -1 | + | (1*1! + 2+2! +…………………………..+ k*k!) +(k+1)*(k+1)! = (k+1)+1! -1 |
| + | = (k+2)! -1 | ||
| − | + | Is it correct? | |
| − | + | ||
| − | Is | + | |
Revision as of 06:50, 21 January 2009
I don't even know how to solve the first problem.
I tried it this way.
Base step: P(0)
0*0 = 0 (0+1)! -1 =0
It is true
Inductive step: Assume P(k) is true then P(k+1) is also true.
P(k) = 1*1! + 2+2! +…………………………..+k*k! = (k+1)! -1 P(k+1) = 1*1! + 2+2! +…………………………..+(k+1)*(k+1)! = (k+2)!-1
From the first equation
(1*1! + 2+2! +…………………………..+ k*k!) +(k+1)*(k+1)! = (k+1)+1! -1
= (k+2)! -1
Is it correct?
