Mathematical Induction

Math Use 33+44+55=?
Acid Outlying ?_?_?_?
Nuclear Element Invisble

1.
let S(n) be the desired statement
for n=1,
LHS = 1/2 = 2 - (3/2) =RHS
hence S(1) is true.
assume S(k) is true
for n=k+1,
1/2 + 2/(2²) + 3/(2³) + ... + k/(2^k) + (k+1)/[2^(k+1)]
= 2 - (k+2)/(2^k) + (k+1)/[2^(k+1)] (by induction hypothesis)
= 2 - (2k+4)/[2^(k+1)] + (k+1)/[2^(k+1)]
= 2 - (2k+4-k-1)/[2^(k+1)]
= 2 - (k+3)/[2^(k+1)]
so S(k+1) is also true
hence S(n) is true for all positive integers n by MI
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2.
let S(n) be the desired statement
for n =1 ,
LHS = 1/(a-1) - 1/a = [a-(a-1)]/[a(a-1)] = 1/[a(a-1)] = RHS
hence S(1) is true.
assume S(k) is true
for n=k+1
1/(a-1) - 1/a - 1/a² -...-1/(a^k) - 1/[a^(k+1)]
= 1/[(a^k)(a-1)] - 1/[a^(k+1)] (by induction hypothesis)
= a/[[a^(k+1)[(a-1)] - (a-1)/[a^(k+1)(a-1)] (since a≠0 and a≠1)
= 1/[[a^(k+1)[(a-1)]
so S(k+1) is also true
hence S(n) is true for all positive integers n by MI
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1.When n = 1,
LHS = 1/2
RHS = 2-3/2
= 1/2
Therefore,the proposition is true for n =1
Assume n = k is true
i.e. 1/2+2/2^(2)+3/2^(3)+...+k/2^(k)=2-(k+2)/2^(k)
When n = k+1
LHS = 1/2+2/2^(2)+3/2^(3)+...+k/2^(k)+(k+1)/2^(k+1)
= 2-(k+2)/2^(k) +(k+1)/2^(k+1)
= 2 -2(k+2)/2^(k+1) + (k+1)/2^(k+1)
= 2 + (- k-3)/2^(k+1)
= 2 - (k+3)/2^(k+1)
= RHS
The proposition is also true for n = k+1
By the principle of MI,1/2+2/2^(2)+3/2^(3)+...+n/2^(n)=2-(n+2)/2^(n) is true
for all positive integers n.

2.When n = 1
LHS = 1/(a-1)- 1/a = 1/a(a-1)
RHS = 1/[a(a-1)]
= LHS
The proposition is true for n =1
Assume the proposition is true for n = k
i.e.1/(a-1)-1/a-1/a^(2)-...-1/a^(k)=1/[a^(k)(a-1)]
When n = k+1
LHS = 1/(a-1)-1/a-1/a^(2)-...-1/a^(k) - 1/a^(k+1)
= 1/[a^(k)(a-1)] - 1/a^(k+1)
= [a - (a - 1)] / [a^(k+1)] (a-1)
= 1 / [a^(k+1)] (a-1)
= RHS
The proposition is also true for n = k+1
By the principle of MI,1/(a-1)-1/a-1/a^(2)-...-1/a^(n)=1/[[a^(n)(a-1)] is true
for all positive integers n,where a =/= 0 and a =/= 1

(唔知有無錯的)