數學歸納法

(a)When n = 1,
(1/6) n (n+1) (2n+1) = (1/6) (1) (2) (3) = 1,
therefore the formula is true for n = 1;

For n + 1
(1/6) n (n+1) (2n+1) + (n+1)^2
= (1/6) (n^2 +n) (2n+1) + (n^2+2n+1)
= (1/6) (2n^3+3n^2+n) + (1/6) (6n^2+12n+1)
= (1/6) (2n^3+9n^2+13n+6)
= (1/6) (n+1) (2n^3+7n+6)
= (1/6) (n+1) (n+2)(2n+3)
= (1/6) (n+1) [(n+1)+1][2(n+1) +1]
Therefore the formula is ture for n+1 while it is true for n.
(b)
1*2+2*3+...+n(n+1)
=Σk(k+1) [k from 1 to n]
=Σ(k^2+k)
=Σk^2+Σk
=1/6n(n+1)(2n+1)+n(n+1)/2
=1/6n(n+1)(2n+1+3)
=1/6n(n+1)(2n+4)
=1/3n(n+1)(n+2)

a)1^2+2^2+...+n^2=1/6n(n+1)(2n+1)

Let P(n) be the proposition "1^2+2^2+...+n^2=1/6n(n+1)(2n+1) "

when n=1,

LHS=1
RHS=(1/6) (1) (1+1) [2(1)+1]=1=LHS

Therefore, P(1) is true.

Assume P(k) is true for some positive integer k.
i.e. 1^2+2^2+...+k^2=1/6k(k+1)(2k+1)

Consider n=k+1,

LHS=1^2+2^2+...+k^2+(k+1)^2
=1/6k(k+1)(2k+1)+(k+1)^2
=1/6(k+1)[(k)(2k+1)+6(k+1)]
=1/6(k+1)[2k^2+7k+6]
=1/6(k+1)[(k+2)(2k+3)]
=1/6(k+1)[(k+1+1)][2(k+1)+1]
=RHS

Therefore,
P(k+1) is also true.

By Principle of Mathematical Induction,
P(n) is true for all positive integer n.

b) 1*2+2*3+...+n(n+1)
=1(1+1)+2(2+1)+...n(n+1)
=1^2+(1)+2^2+(2)+...n^2+(n)
=(1^2+2^2+...n^2)+(1+2+...n)
=1/6n(n+1)(2n+1)+[(n)(1+n)/2] by a)
=n(n+1)[1/6(2n+1)+1/2]
=n(n+1)[1/6(2n+4)]
=1/3n(n+1)(n+2) //

我用英文可以嗎？
另外(a)打錯了題目喔(^^")

(a)
Put n=1, LHS=1^2=1,RHS=(1/6)*(1)(2)(3)=1,Thus,the proposition is true for n=1.
Assume that the proposition is true for n=k,
i.e.1^2+2^2+...+k^2=(1/6)*(k)(k+1)(2k+1)
Let n=k+1,
Then, 1^2+2^2+...+k^2+(k+1)^2
=(1/6)(k)(k+1)(2k+1)+(k+1)^2
=(k+1)((2(k)^2+k)/6+(k+1))
=(k+1)(1/6)(2(k)^2+k+6k+6)
=(1/6)(k+1)(2(k)^2+7k+6)
=(1/6)(k+1)(k+2)(2k+3)
=(1/6)(k+1)(k+2)(2(k+1)+1)
Thus,the proposition is true for n=k+1.
The proposition is true for any positive n.

(b)
1*2+2*3+...+n(n+1)
=1*(1+1)+2*(2+1)+...+n(n+1)
=1*1+2*2+...+n*n+1+2+...+n
=1^2+2^2+...+n^2+1+2+...+n
by part(a),
=(1/6)(n)(n+1)(2n+1)+(n+1)(n)/2
=n(n+1)(2n+1)/6+3/6
=n(n+1)(n+2)/3