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[Nawreez]

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[محشش الفيلسوف]

شـــــــــــــــــــــــــــــــــــــــــــــــــ ــــــــــكراً

[Nawreez]

Problem 1:

Use mathematical induction to prove that

1 + 2 + 3 + ... + n = n (n + 1) / 2

for all positive integers n.

Solution to Problem 1:

• Let the statement P (n) be
  
  1 + 2 + 3 + ... + n = n (n + 1) / 2
• STEP 1: We first show that p (1) is true.
  
  Left Side = 1
  
  Right Side = 1 (1 + 1) / 2 = 1
• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true
  
  1 + 2 + 3 + ... + k = k (k + 1) / 2
• and show that p (k + 1) is true by adding k + 1 to both sides of the above statement
  
  1 + 2 + 3 + ... + k + (k + 1) = k (k + 1) / 2 + (k + 1)
  
  = (k + 1)(k / 2 + 1)
  
  = (k + 1)(k + 2) / 2
• The last statement may be written as
  
  1 + 2 + 3 + ... + k + (k + 1) = (k + 1)(k + 2) / 2
• Which is the statement p(k + 1).

Problem 2:

Prove that

1 2 + 2 2 + 3 2 + ... + n 2 = n (n + 1) (2n + 1)/ 6

For all positive integers n.

Solution to Problem 2:

• Statement P (n) is defined by
  
  1 2 + 2 2 + 3 2 + ... + n 2 = n (n + 1) (2n + 1)/ 2
• STEP 1: We first show that p (1) is true.
  
  Left Side = 1 2 = 1
  
  Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1
• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true
  
  1 2 + 2 2 + 3 2 + ... + k 2 = k (k + 1) (2k + 1)/ 6
• and show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statement
  
  1 2 + 2 2 + 3 2 + ... + k 2 + (k + 1) 2 = k (k + 1) (2k + 1)/ 6 + (k + 1) 2
• Set common denominator and factor k + 1 on the right side
  
  = (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6
• Expand k (2k + 1)+ 6 (k + 1)
  
  = (k + 1) [ 2k 2 + 7k + 6 ] /6
• Now factor 2k 2 + 7k + 6.
  
  = (k + 1) [ (k + 2) (2k + 3) ] /6
• We have started from the statement P(k) and have shown that
  
  1 2 + 2 2 + 3 2 + ... + k 2 + (k + 1) 2 = (k + 1) [ (k + 2) (2k + 3) ] /6
• Which is the statement P(k + 1).

Problem 3:

Use mathematical induction to prove that

1 3 + 2 3 + 3 3 + ... + n 3 = n 2 (n + 1) 2 / 4

for all positive integers n.

Solution to Problem 3:

• Statement P (n) is defined by
  
  1 3 + 2 3 + 3 3 + ... + n 3 = n 2 (n + 1) 2 / 4
• STEP 1: We first show that p (1) is true.
  
  Left Side = 1 3 = 1
  
  Right Side = 1 2 (1 + 1) 2 / 4 = 1
• hence p (1) is true.
• STEP 2: We now assume that p (k) is true
  
  1 3 + 2 3 + 3 3 + ... + k 3 = k 2 (k + 1) 2 / 4
• add (k + 1) 3 to both sides
  
  1 3 + 2 3 + 3 3 + ... + k 3 + (k + 1) 3 = k 2 (k + 1) 2 / 4 + (k + 1) 3
• factor (k + 1) 2 on the right side
  
  = (k + 1) 2 [ k 2 / 4 + (k + 1) ]
• set to common denominator and group
  
  = (k + 1) 2 [ k 2 + 4 k + 4 ] / 4
  
  = (k + 1) 2 [ (k + 2) 2 ] / 4
• We have started from the statement P(k) and have shown that
  
  1 3 + 2 3 + 3 3 + ... + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4
• Which is the statement P(k + 1).

Problem 4:

Prove that for any positive integer number n , n 3 + 2 n is divisible by 3

Solution to Problem 4:

• Statement P (n) is defined by
  
  n 3 + 2 n is divisible by 3
• STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n
  
  1 3 + 2(1) = 3
  
  3 is divisible by 3
• hence p (1) is true.
• STEP 2: We now assume that p (k) is true
  
  k 3 + 2 k is divisible by 3
  
  is equivalent to
  
  k 3 + 2 k = 3 M , where M is a positive integer.
• We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms
  
  (k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3
  
  = [ k 3 + 2 k] + [3 k 2 + 3 k + 3]
  
  = 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ]
• Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.

[Nawreez]

Example 1: Prove by the principle of mathematical induction method 4 + 8 + 12 + … + 4n= 2n(n+ 1)
Solution: P(n)= 4 + 8 + 12 + … + 4n= 2n(n + 1)
Put n= 1
P(1) true
Assume that statement be true for n= k
Assume P(k) true.
Assume 4 + 8 + 12 + … + 4k= 2k(k+ 1) be true
To prove P(k+ 1) - true.
4 + 8 + 12 + … + 4(k+1)= 2(k+1)(k+1+ 1)= 2(k+1)(k+2)
P(k+ 1) is true.
If P(k) true, then P(k+ 1) is also true.
P(n) is true for all n ∈ N.

^
ملاحظة على المثال ,,

بعض الطلبة . يستعملون طريقة معينة ف التبسيط بحيث يحذفون الرقم الا يطلعونه ,, مثلا 2 .. بعضهم مايخلونه في الجواب النهائي ..
وهذا غلط ,, عشان جذي تاكدوااا انه جوابكم النهائي يطابق تمامااا أول معطى في السؤال ,,

[Nawreez]

Example: Adding up Odd Numbers

1 + 3 + 5 + ... + (2n-1) = n2

1. Show it is true for n=1

1 = 12 is True

2. Assume it is true for n=k

1 + 3 + 5 + ... + (2k-1) = k2 is True

Now, prove it is true for "k+1"

1 + 3 + 5 + ... + (2k-1) + (2(k+1)-1) = (k+1)2 ... ?

We know that 1 + 3 + 5 + ... + (2k-1) = k2 (the assumption above), so we can do a replacement for all but the last term:

k2 + (2(k+1)-1) = (k+1)2

Now expand all terms:

k2 + 2k + 2 - 1 = k2 + 2k+1

And simplify:

k2 + 2k + 1 = k2 + 2k + 1
They are the same! So it is true.

So:

1 + 3 + 5 + ... + (2(k+1)-1) = (k+1)2 is True

DONE!