it is very simple!!!!!!!!!
1.1!+2.2!+............+k.k!+(k+1).(k+1)!={(k+1)!-1}+(k+1).(k+1)!
=(k+1)!.(k+2)-1
=(k+2)!-1
hence the proof
When doing problems that tends to these 3 steps: Basis Step, Inductive Hypothesis and proving the Inductive Step, I am having difficuly understanding the Inductive Hypotheses. How do you know what to assume for the Hypothesis? This is what I have.
Q: Prove that 1 · 1! + 2 · 2! + ··· n · n! = (n + 1)! – 1 whenever n is a positive integer?
I have this:
Basis step:
Let n = 1
n · n! = (n + 1)! – 1
1 · 1! = (1 + 1)! – 1
1 · 1 = (2)! – 1
1 = 1
Since they both equal, the basis step holds.
Inductive Hypothesis:
1 · 1! + 2 · 2! + ··· (k + 1) · (k + 1)! = ((k + 1) + 1)! – 1
Would I add a (k + 2)! to both the LHS and RHS?
Step 3: Show that it follows from step 2 that 1 · 1! + 2 · 2! + ··· (k + 1) · (k + 1)! = ((k + 1) + 1)! – 1.
1 · 1! + 2 · 2! + ··· k · k! + (k + 1) · (k + 1)! = (k + 1)! - 1 + (k + 1) · (k + 1)! = (k + 1)! [1 + (k + 1)] - 1 = (k + 1)! [(k + 1) + 1] - 1 = ((k + 1) + 1)! – 1
and the job is almost done. Now you dot the i's and cross the t's by making the final statement about having proved the result.
The "inductive hypothesis" is simply that whatever your theorem is, it is true for some specific number, k, say. Notice that the theorem says that it is true for ALL numbers while the inductive hypothesis is only that it is true for SOME number.
It would be better to write just exactly what your equation is with "k" rather than "n": 1 · 1! + 2 · 2! + ··· k ·k! = (k + 1)! – 1Q: Prove that 1 · 1! + 2 · 2! + ··· n · n! = (n + 1)! – 1 whenever n is a positive integer?
I have this:
Basis step:
Let n = 1
n · n! = (n + 1)! – 1
1 · 1! = (1 + 1)! – 1
1 · 1 = (2)! – 1
1 = 1
Since they both equal, the basis step holds.
Inductive Hypothesis:
1 · 1! + 2 · 2! + ··· (k + 1) · (k + 1)! = ((k + 1) + 1)! – 1
No. You write the left hand side of the original equation, 1 · 1! + 2 · 2! + ··· n · n!, with k+1 rather than k- which means you would add (k+1)(k+1)! to both sides:Would I add a (k + 2)! to both the LHS and RHS?
1· 1!+ 2· 2!+ ···+ k·k!+ (k+1)·(k+1)!= (k+1)!- 1+ (k+1)(k+1)!
You want to make the right side look like ((k+1)+ 1)!- 1. I suggest that you factor (k+1)! out of two parts of you previous equation.