Question

PLZ SOLVE THIS PROBLEM



Prove by the principle of mathematical induction that for all n€N:
(i)1+3+5 + 7 +.....+(2n-1) = n2
(ii)2 + 4 + 6 + 8 +.... +2n = n(n + 1)
(iii)1 +2 +3 +4 +......+n=(1/2)n(n + 1).​

Answer 1

Answer:

1. P (1) = 1

RHS = 1^2 = 1

P(1) is true.

P(k) = 1+3+5 + 7 +.....+(2k-1) = k^2

P ( k +1) = 1+3+5 + 7 +.....+(2k-1)+2( k +1) =( k+1)^2

=> k^2 + 2(k+1) = k^2 + 2k +1

=> ( k + 1) ^2 = R. H. S

BY, PMI, p( k + 1 ) is also true.

(2 ). P(1) = 2

RHS = 1( 1 +1) = 2

P(1)is true.

P ( k) = 2 + 4 + 6 + 8 +.... +2k = k(k + 1)

P(k+1) = 2+4+6+8+...+2k + 2(k+1) = (k+1){(k+1)+1}

=> 2+4+6+8+...+2k + 2(k+1) = (k+1)(k+2)

=> k(k+1) + 2( k + 1)

=> (k+1)(k+2) = RHS

By PMI, P(k+1) is also true .

( 3). P(1) = 1

RHS = 1/2 ( 1 +1) = 2/2 = 1

P(1) is true. Bcz LHS = RHS = 1

P(k) = 1 +2 +3 +4.. + k =( 1/2) k(k+1)

P(k+1) = 1 + 2 + 3 +4 +... +k + ( k+1) = 1/2 ( k+1){(k+1)+1} = (k+1)(k+2)/2

LHS. = k(k+1)/2 + ( k+ 1 ) = {k(k+1) + 2( k+1)} /2

=>(k+1)(k+2)/2 = RHS

By PMI, P( k+1) is also true.

♥️QED