F.4 AM

Let P ( n ) be the proposition
“(1/2)+(3/22)+(5/23)+....+(2n-1/2n)=3-(2n+3/2n)”.
When n = 1,
L.H.S. = ( 2 – 1 ) / 21 = 1 / 2
R.H.S. = 3 – ( 2 + 3 ) / 21 = 1 / 2 = L.H.S.
So P ( 1 ) is true.
Assume P ( k ) is true for some positive integers k, i.e.
(1/2)+(3/22)+(5/23)+....+(2k-1/2k)=3-(2k+3/2k)
When n = k + 1,
L.H.S. = (1/2)+(3/22)+(5/23)+....+(2k-1/2k) + (2k+2-1/2k+1)
= 3-(2k+3/2k) + (2k+2-1/2k+1)
= 3 –(2k+3/2k) + (2k+1/2k )( 1 / 2 )
= 3 + ( 1 / 2 )[ (2k+1)/2k – (4k+6)/2k ]
= 3 + ( 1 / 2 x 2k )( 2k + 1 – 4k – 6 )
= 3 + ( 1 / 2k+1 )( -2k – 5 )
= 3 – ( 2k + 5 ) / 2k+1
R.H.S. = 3 – ( 2k + 2 + 3 ) / 2k+1
= 3 – ( 2k + 5 ) / 2k+1
= 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)nC2 = 55
n! / 2! ( n – 2 )! = 55
( 1 / 2 )( n )( n – 1 ) = 55
n2 – n = 110
n2 – n – 110 = 0
( n – 11 )( n + 10 ) = 0
n = 11 or n = -10 ( rejected )
Hence n = 11.

1)
for n = 1
LHS = 1/2
RHS = 3-(2+3)/2 = 1/2
Therefore, the equation is true for n = 1

Assume the equation is true for n = k
ie. (1/2) + (3/2^2) + ... + (2k-1)/2^k = 3 - (2k+3)/2^k
When n = k+1
LHS = (1/2) + (3/2^2) + ... + (2k-1)/2^k + [2(k+1)-1]/2^(k+1)
= 3 - (2k+3)/2^k + [2(k+1)-1]/2^(k+1)
= 3 - (4k+6)/2^(k+1) + [(2k+1)]/2^(k+1)
= 3 - [(4k+6) - 2k - 1]/2^(k+1)
= 3 - (2k+5)/2^(k+1)
= 3 - [2(k+1)+3]/2^(k+1)
= RHS

Therefore, the equation is true for all positive integer n according to MI

2)

nC2
= n (n-1)/2
= 55

n^2 - n = 110
n^ - n - 110 = 0
(n-11)(n+10) = 0
n = 11 or n = -10 (rejected)

Therefore, n = 11

諗左好耐先知你打緊咩
-(2n+3/2^n) 應該係 [(2n+3)/(2^n)] 記住先成除後加減
Let P(n) be the proposition '(1/2)+(3/2^2)+(5/2^3)+....+[(2n-1)/2^n]=3-[(2n+3)/2^n]'
when n=1
LHS = 1/2
RHS = 3-(2+3)/2 = 1/2
so P(1) is true
Assume that P(k) is true for some positive integer k
i.e. (1/2)+(3/2^2)+(5/2^3)+....+ [(2k-1)/2^k]=3-[(2k+3)/2^k]
When n=k+1
LHS
=(1/2)+(3/2^2)+(5/2^3)+....+[(2k-1)/2^k] + [2(k+1)-1]/[2^(k+1)]
=3-[(2k+3)/2^k] + (2k+1)/[2^(k+1)]
=3- (4k + 6) / [(2^k) x 2] +(2k+1)/[2^(k+1)]
=3- (4k + 6) / [2^(k+1)] +(2k+1)/[2^(k+1)]
=3 + (-4k - 6 + 2k +1) / [2^(k+1)]
=3+ (-2k-5)/ [2^(k+1)]
=3-(2k+5) / [2^(k+1)]

RHS
=3- [2(k+1)+3]/[2^(k+1)]
=3- (2k+5) / [2^(k+1)]
=LHS
so P(k+1) is true
By the principle of mathematical induction
P(n) is true for all positive integers n

2)
nC2 = 55
n(n-1) / 2 = 55
n^2 - n = 110
n^2 - n -110 =0
(n-11)(n+10) = 0
n = 11 or -10 (rejected as n cannot be negative)
so n =11