a-maths mathematical induction

a) Prove by mathematical induction that 1^(2)+2^(2)+3^(2)+...+n^(2)=1/6．ｎ(n+1)(2n+1)for all positive integers n.

(a)
let S(n) be the statement " 1^(2)+2^(2)+3^(2)+...+n^(2)=1/6．ｎ(n+1)(2n+1)for all positive integers n."

when n=1
LHS=1
RHS=1

so, when n=1 the statement is true

Assume when n=k, the statement is true that is
1^(2)+2^(2)+3^(2)+...+k^(2)=1/6*k(k+1)(2k+1) for all positive integers n.

when n=k+1
1^(2)+2^(2)+3^(2)+...+k^(2)+(k+1)^2
=1/6*k(k+1)(2k+1)+(k +1)^2
=1/6(k+1)[k(2k+1)+6k+6]
=1/6(k+1)[2k^2+7k+6]
=1/6(k+1)(k+2)(2k+3)

when n=k+1, the statement is true
By MI for all values of n, S(n) is true

b)
(ii)
2^(2)+4^(2)+6^(2)+...+(2m)^(2)
=(2^2)(1^2)+(2^2)(2^2)+(2^2)(3^2)+...+(2^2)(m^2)
=(2^2)[(1^2)+(2^2)+(3^2)+...+(m^2)]
=4*1/6(m)(m+1)(2m+1 )
=2/3(m)(m+1)(2m+1)
(iii)
1^2 + 3^2 + 5^2 +...+ (2m)^2
=[1^2+2^2+3^2+....+(2m)^2]-[2^2 + 4^2 + 6^2 + ... +(2m)^2]
=1/6 (2m)( 2m+1)( 4m+1)-2/3(m)(m+1)(2m+1)
=1/3(m)( 2m+1)( 4m+1)-2/3(m)(m+1)(2m+1)
=1/3(m)(2m+1)[4m+1-2m-2]
=1/3(m)(2m+1)(2m-1)





2007-01-07 22:52:44 補充：
最後個個表達式應該是1^2 + 3^2 + 5^2 +...+ (2m-1)^2