數學歸納法.......

(1) Let P(n) be the proposition
“2+5+8+.....+(3n-1)=n (3n+1)/2”.
When n = 1,
L.H.S. = 3 – 1 = 2
R.H.S. = (1)(3+1)/2 = 2 = L.H.S. So P (1) is true.
Assume P(k) is true for some positive integers k, i.e.
2+5+8+.....+(3k-1)=k(3k+1)/2
When n = k+1,
L.H.S. = 2+5+8+…+(3k-1)+(3k+3-1)
= k(3k+1)/2 +(3k+2)
= [k(3k+1)+2(3k+2)]/2
= (3k2 + k + 6k + 4 )/2
= (3k+4)(k+1)/2
R.H.S. = (k+1)(3k+3+1)/2
= (k+1)(3k+4)/2
=L.H.S. So P ( k + 1 ) is true.
By the principle of mathematical induction, P(n) is true for all positive integers n.
(2) Let P(n) be the proposition
“3+32+33+.....+3n=3(3n-1)/2”.
When n = 1,
L.H.S. = 31 = 3
R.H.S. = 3(31 – 1 )/2 = 3 = L.H.S. So P (1) is true.
Assume P(k) is true for some positive integers k, i.e.
3+32+33+.....+3k=3(3k-1)/2
When n = k + 1,
L.H.S. = 3+32+33+.....+3k+3k+1
=3(3k-1)/2 + 3k+1
= [3(3k-1)+2(3k+1)] / 2
= [3k+1 – 3 + 2(3k+1)] / 2
= 3[3k+1-1]/2
R.H.S. = 3(3k+1-1)/2
=L.H.S. So P ( k + 1 ) is true.
By the principle of mathematical induction, P(n) is true for all positive integers n.

let P(n) be the statement"2+5+8+.....+(3n-1)=n(3n+1)/2"
when n=1,
LHS=2
RHS=2
P(1) is true
assume P(k) is true for any positive integer k
i.e"2+5+8+.....+(3k-1)=k(3k+1)/2"
when n=k+1
RHS=(k+1)(3k+4)/2
LHS=2+5+8+.....+(3k-1)+(3k+3-1)
......=k(3k+1)/2+3k+2
......=(3k^2+k+6k+4)/2
......=(k+1)(3k+4)/2
so,P(k+1) is also true if P(k) is true for any positive integer k
by principle of M.I ,P(n) is true for all positive integers n


let P(n) be the statement"3+3^2+3^3+.....+3^n=3(3^n-1)/2"
when n=1,
LHS=3
RHS=3
P(1) is true
assume P(b) is true for any positive integer b
i.e 3+3^2+3^3+.....+3^b=3(3^b-1)/2
when n=b+1
RHS=3(3^(b+1)-1)/2
.......=(9(3)^b-3)/2
LHS=3+3^2+3^3+.....+3^b+3^(b+1)
......=3(3^b-1)/2+3^(b+1)
......=(3(3)^b-3+2(3)(3)^b)/2
......=(3(3)^b(1+2)-3)/2
......=(9(3)^b-3)/2
so,P(b+1) is also true if P(b) is true for any positive integer b
by principle of M.I ,P(n) is true for all positive integers n

1.
Let P(n) be the proposition of the statement "2+5+8+.....+(3n-1)=n(3n+1)/2"
When n=1, LS = 3(1)-1 = 2 RS = 1(4)/2 = 2
Therefore LS = RS
P(1) is true
Assume P(k) is true
i.e. 2+5+8+...+(3k-1) = k(3k+1)/2
then for n=k+1
LS = 2+5+8+...+(3k-1) + [3(k+1)-1]
= k(3k+1)/2 + (3k+2)
= (3k^2+k)/2 + (6k+4)/2
= (3k^2+7k+4)/2
= (k+1)(3k+4)/2
= (k+1)[3(k+1)+1]/2 = RS
Therefore, P(k+1) is also true
By MI, P(n) is true for all integers n.

2.
Let P(n) be the proposition of the statement "3+3^2+3^3+.....+3^n=3(3^n-1)/2 "
When n=1, LS = 1 RS = (3-1)/2 = 1
Therefore LS = RS and P(1) is true
Assume P(k) is true
i.e. 3+3^2+3^3+.....+3^k=3(3^k-1)/2
then for n = k+1
LS = 3+3^2+3^3+.....+3^k + 3^(k+1)
= 3(3^k-1)/2 + 3^(k+1)
= (3^k*3 - 3)/2 + [2*3^k(3)]/2
= {3^k*3 - 3 + [2*3^k(3)]} / 2
= [3^(k+1) - 3 + 2*3^(k+1)] / 2
= 3[3^k - 1 + 2*3^k] / 2
= 3[3^k(1+2) -1] / 2
= 3[3^(k+1) -1] / 2 = RS
Therefore, P(k+1) is also true
By MI, P(n) is true for all integers n.