Proof by Induction

 

1+2+3+...+n=12n(n+1)

 

STEP 1

For n = 1:

1=12(1)(1+1)1=1LHS=RHS

The statement is true for n = 1


STEP 2

Assume the statement is true for n = k:

1+2+3+...+k=12k(k+1)

 

So, for n = k + 1:

1+2+3+...+k+(k+1)=12(k+1)(k+2)

 

LHS1+2+3+...+k+(k+1)1+2+3+...+k=12k(k+1)12k(k+1)+(k+1)(k+1)(12k+1)12(k+1)(k+2)LHS=RHS

If the statement is true for n = k, then it is true for n = k + 1


STEP 3

Based on step 1 and 2, then:

The statement is true for n = 1, then the statement is true for n = 2

The statement is true for n = 2, then the statement is true for n = 3

and so on,

The statement is true for n ≥ 1

Exercise

Proof the statements below by mathematical induction

 

1+8+15++(7n6)=12(7n25n)

11!.3+12!.4+13!.5+...+1n!(n+2)=121(n+2)!

r=1n3r=32(3n1)

r=1n2r13r=1n+13n

5n+3 is divisible by 4, for n1

22n+1+32n+1 is divisible by 5, for n1

3n>1+2n, for n>1

(x+y)n>xn+yn, untuk n>1

sinx+sin2x+sin3x+...+sinnx=sin12(n+1)xsin12nxsin12x

cosx+cos2x+cos3x+...+cosnx=cos12(n+1)xsin12nxsin12x

Proof by Induction