Mathematical induction666

Prove, by mathematical induction, that for all positvie integers n, 1+2+3+4+...+2^(n-1) = 2^(2n-3) + 2^(n-2).

When n=1 ,

1+...+2^(1-1)=1=2^(2-3)+2^(1-2) , it is true for n=1 ,

Assume it is true for n =k , where k is a positive integer ,

i.e.1+2+3+...+2^(k-1)=2^(2k-3)+2^(k-2)

Consider n=k+1 ,

1+2+3+...+2^(k-1)+[2^(k-1)+1]...+2^(k) <--- always get wrong there

=2^(2k-3)+2^(k-2)+...+2^(k)

=2^(2k-3)+2^(k-2)+2^(k-1)x[2^(k-1)+1]+(1/2)[2^(k-1)+1](2^k)+2^(k)

=2^(2k-3)+2^(k-1)+2^(k-1)[2^(k-1)+1+2^(k-1)+1]+2^(k)

=2^(2k-3)+2^(k-1)+2^(k-1)[2^(k)+2]+2^(k)

=2^(2k-3)+2^(2k-1)+2^(k-1)+2^(k-1)+2^k

=2^(2k-2)+2^(k-1) , it is true for n = k+1 ,

By mathermatical induction , it is true for all positive integer n.




2009-10-01 20:40:24 補充：
2/(x-2) - [(2/x)+(2^2/x^2) +...+ (2^n/x^n)] = 2^(n+1) / [(x^n) (x-2)].

i do the " assume " and "consider" parts only

Assume it is true for n=k , where k is a positive integer ,

i.e. 2/(x-2) - [(2/x)+(2^2/x^2) +...+ (2^k/x^k)] = 2^(k+1) / [(x^k) (x-2)].

2009-10-01 20:42:58 補充：
Consider n=k+1 ,

2/(x-2) - [(2/x)+(2^2/x^2) +...+ (2^k/x^k)+2^(k+1)/x^(k+1)]

=2^(k+1) / [(x^k) (x-2)]-2^(k+1)/x^(k+1)

=(2^(k+1)/(x^k))[1/(x-2)-1/x]

=(2^(k+1)/(x^k))(x-x+2)/x(x-2)

=2^(k+2)/x^(k+1)(x-2) it is true for n=k+1 ,

By Mathematical induction , it is true for all positive integer n.