AMATHS.......A LITTLE HELP, PLX~~~~

1.let P(n) be the statement"1+2x2+3x2^2+4x2^2+......+n x 2^ (n-1) = 1+( n-1) 2^n"
when n=1
LHS=1
RHS=1
P(1)is true
assume P(k) is true for any positive integer k
ie "1+2x2+3x2^2+4x2^2+......+ k x2^ (k-1) = 1+( k-1) 2^k
when n=k+1
RHS=1+(k)2^(k+1)
LHS=1+2x2+3x2^2+4x2^2+......+ k x2^ (k-1)+(k+1)2^k
.......=1+( k-1) 2^k+(k+1)2^k
.......=1+(k)2^k-2^k+(k)2^k+2^k
.......=1+(2k)2^k
.......=1+(k)2^(k+1)
so P(k+1) is true if P(k) is true for any positive integer k
By principle of MI,P(n) is true for all positive integers n


2.let P(n) be the statement"1+ 2^1 +2^2 +.......+2^n = 2^(n+1) -1"
when n=0
LHS=1
RHS=1
P(0)is true
when n=1.......(*)
LHS=1+2=3
RHS=3
P(1)is true
assume P(k) is true for any positive integer k
ie "1+ 2^1 +2^2 +.......+2^k = 2^(k+1) -1"
when n=k+1
RHS=2^(k+2) -1
LHS=1+ 2^1 +2^2 +.......+2^k+2^(k+1)
.......=2^(k+1) -1+2^(k+1)
.......=2^(k+1)(1+1)-1
.......=2^(k+2)-1
so P(k+1) is true if P(k) is true for any positive integer k
By principle of MI,P(n) is true for all positive integers n and n=0


3.result(1)+result(2)......(**)
LHS=(1+2x2+3x2^2+4x2^2+......+n x 2^ (n-1))+(1+ 2^1 +2^2 +.......+2^(n-1) )
......=2+2(2+1)+(2^2)(3+1)+(2^3)(4+1)......+(n+1)2^(n-1)

RHS=1+( n-1) 2^n+2^(n+1-1) -1
.......=(n)2^n-2^n+2^n
.......=(n)2^n
From result (1)and(2),we know LHS=RHS
so,2+2(2+1)+(2^2)(3+1)+(2^3)(4+1)......+(n+1)2^(n-1)=(n)2^n
put n=99,
2+2(3)+(4)2^2+(5)2^3......+(100)2^(98)=(99)2^99

(*)第2條的式係由n=0開始....你可由n=0開始證,so 佢係由0至所有positive integers都成立
而第1條係由n=1開始成立....所以你唔能夠由n=0開始證
(**)在第三題中,(1)+(2),因為要balance條式中的n值
因此當(1)中的n=1時,係=(2)中的n=1-1

1. when n=1
LHS : 1 * 2^(1-1) = 1
RHS : 1 + (1-1) 2^1 = 1
LHS=RHS
so the proporsition is true for n= 1
assume the proporsition is true for n=k
ie. 1+2x2+3x2^2+4x2^2+......+k x 2^ (k-1) = 1+( k-1) 2^k
when n=k+1
1+2x2+3x2^2+4x2^2+......+k x 2^ (k-1) + (k+1) x 2^ ((k+1)-1)
=1+( k-1) 2^k + (k+1) x 2^ ((k+1)-1)
=1+ k 2^k - 2^k + (k+1)2^k
=1 + k 2^k +k 2^k -2^k+2^k
=1 + 2k 2^k
=1+k 2^k+1
hence the result.