mathematical induction 2

我想你的問題有少少打錯.. 應該是 1/2 + 2/2^2 + 3/2^3 +．．．+ n/2^n = 2 - (n+2)/2^n

Let P(n) be 1/2 + 2/2^2 + 3/2^3 +．．．+ n/2^n = 2 - (n+2)/2^n for all positive integers n.

When n = 1,

Left side = 1/2^1
= 1/2

Right side
= 2-(1+2)/2^1
= 2-3/2
= 1/2

As right side = left side, so P(1) is true.

Assume P(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,

Left side
= 1/2 + 2/2^2 + 3/2^3 +．．．+ k/2^k + (k+1)/2^(k+1)

= (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) ...................... by assumsion of P(k)

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

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

= (1/2^(k+1)) [2^(k+2) - 2k -4 + k + 1]

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

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

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

= Right side

So P(k+1) is true.

By Mathematical Induction, 1/2 + 2/2^2 + 3/2^3 +．．．+ n/2^n = 2 - (n+2)/2^n is true for all positive integers n.

Let S(n) be the statement 1/2+2/2^2+3/2^3+......+n/2^n=2-(n+2)/2^n
When n=1
L.H.S.=1/2
R.H.S.=2-(1+2)/2^1=1/2
Therefore S(1) is true.
Let S(k) is true.
When n=k+1
L.H.S.=2-(k+2)/2^k+(k+1)/2^(k+1)
=(k+1)/2^(k+1)-2(k+2)/2^(k+1)+2
=[(k+1)-2(k+2)]/2^(k+1)+2
=(-k-3)/2^(k+1)+2
=2-(k+3)/2^(k+1)
R.H.S.=2-(k+1+2)/2^(k+1)
=2-(k+3)/2^(k+1)
Therefore L.H.S.=R.H.S.
Therefore by the principle of M.I.,S(n) is true for all natural number n.

'1/2 + 2/2^2 + 3/2^2 +．．．+ n/2^n = 2 - (n+2)/2^n'
Let P(n) be the statement

When n=1
L.H.S.=1/2
R.H.S.=2-(1+2)/2=1/2
L.H.S.=R.H.S.
P(1) is true

Assume P(k) is true, where k is some positive integers
1/2 + 2/2^2 + 3/2^2 +．．．+ k/2^k = 2 - (k+2)/2^k

When n=k+1
L.H.S.=1/2 + 2/2^2 + 3/2^2 +．．．+ k/2^k + (k+1)/2^(k+1)
=2 - (k+2)/2^k + (k+1)/2^(k+1)
=2 - (k+2)/2^k + (k+1)/(2*2^k)
=2 - (2k+4)/(2*2^k) + (k+1)/(2*2^k)
=2 - (k+3)/(2*2^k)
=2 - ((k+1)+2)/2^(k+1)
R.H.S.=2 - ((k+1)+2)/2^(k+1)
L.H.S.=R.H.S.
P(k+1) is true

By the principle of mathematical induction, P(n) is true for all positive integers n.