Solution (basis step and inductive step):Proof that p(1) is true. When n = 1, the LHS of P(1) equals 1, and the

RHS equals 12 which also equals 1. So P(1) is true:

Proof that for all integers k >= 1, if P(k) is true then P(k + 1) is true: Suppose that k is any integer with k >=1 such

that 1 + 3 + 5 + …. + (2k - 1) = k: [This is the inductive hypothesis, P(k).] We must show that P(k + 1) is true,

where P(k + 1) is the equation1 + 3 + 5 + …..+ (2k + 1) = (k + 1)²

1 + 3 + 5 +....+ (2n - 1) = n²

Now the LHS of P(k + 1) is 1 + 3 + 5 + ….. + (2k + 1) = 1 + 3 + 5 +…. + (2k-1) + (2k + 1) by making the next-to-last term explicit

= k² + (2k + 1) by inductive hypothesis.

And the RHS of P(k + 1) is(k + 1)² = k²+ 2k + 1 by basic algebra.

k² + (2k + 1) = k²+ 2k + 1 So the left-hand and right-hand sides of P(k+1)

equal the same quantity, and thus and thus P(k+1)

is true [as was to be shown].

INDUCTION HYPOTHESISExample :

 4 + 7 + 10 +- - - - + 3n+2= ½(3n² +5n ) prove by inductive.

-assume it is true ( hypothesis ) and then 

If n= n+1 the sum should be ½ [3(n+1)² +5(n+1). prove it

-The term of n+1 = 3(n+1)+2 

-The sum of (n+1) = ½(3n² +5n )+ 3(n+1)+2

½(3n² +5n + 6n+6+4)

= ½ [(3n² +6n + 3) + 5n+5

= ½[3(n+1)²+ 5(n+1)]

  ---> if inductive hypothesis P(n) is true , then P(n+1) must be true

MATHEMATICAL INDUCTION TO CONSTRUCT PROOFS INVOLVING

FORMULAS, DIVISIBILITY PROPERTIES AND INEQUALITIES

FORMULA : SUMMATION FORMULAE 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) / 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).

PROVING INEQUALITIESn<2ⁿFor all positive integer nLet p(n)=n<2ⁿ

Basis step:P(1) is true1<2¹=2

Inductive step: p(k) is true, then p(k+1) is trueP(k)= k<2ᴷP(k+1)= k+1<2ᴷᶧ¹

Add 1 to both sidesK+1 < 2ᴷ+1 <= 2ᴷ+2ᴷ =2.2ᴷ =2ᴷᶧ¹

P(k+1) is true

PROVING DIVISIBILITY Prove n³-n is divisible by 3. n is positve integer

Basis step: p(1) is true1³-1=0 is divisible by 3

Inductive step: p(k) is trueP(k)= k³-k is divisible by 3P(k+1)= (k+1) ³-(k+1) is divisible by 3

P(k+1) ³-(k+1 ) = (k³+3k²+3k+1)- (k+1 ) (k³-k)+3(k²+k)