I am kinda new to proof and can find it difficult sometimes. Can some help me to solves the following proofs.
1. d/dx (x^n)= nx^n-1
2. r(r+1) = 1/3 n(n+1)(n+2)
3. 1.1!+ 2.2! + 3.3! +...+ n.n! = (n+1)!-1
Thanks
These all look like "proof by induction" problems. Prove the statement is true for n= 1 (the first does not say that n is a positive integer but I am assuming that), the prove "if the statement is true for n=k then it is true for n= k+1.
If n= 1, this says d(x^1)/dx= 1(x^0). That is true because y= x is a linear function with slope 1= 1(x^0).1. d/dx (x^n)= nx^n-1
Suppose you know that d/dx (x^k)= k x^k-1. x^k+1= x(x^k). Differentiate that using the product rule.
I don't know what you mean here. Assuming you really mean [tex]r^k(r)(r+1)[/itex] this is NOT true even for n= 1. If you mean $\displaystyle \sum_{k=1}^n k(k+1)$, when n= 1 the left sides is just 1(2)= 2 and the right sides is (1/3)(1)(2)(3)= 2 so it is true for n= 1.2. r(r+1) = 1/3 n(n+1)(n+2)
Suppose it is true for n= i:
$\displaystyle \sum_{k=1}^i k(k+1)= (1/3)i(i+1)(i+2)$
Then [tex]\sum_{k=1}^{i+1} n(n+1)= \sum_{k=1}^i k(k+1) + (i+1)(i+2)[/itex]
When n= 1, the left sides is 1.1!= 1 and the right side is 2!- 1= 1 so it is true for n= 1.3. 1.1!+ 2.2! + 3.3! +...+ n.n! = (n+1)!-1
Assume 1.1!+ 2.2! + 3.3! +...+ k.k! = (k+1)!-1. Then 1.1!+ 2.2! + 3.3! +...+ (k+1).(k+1)! = (k+1)!-1 + (k+1).(k+1)! Try factoring out (k+1)!
Thanks[/QUOTE]