Question

Prove by method of induction, for all n ∈ N
$\rm 1^{2}+3^{2}+5^{2}+... +(2n-1)^{2} = \frac{n}{3}(2n-1)(2n+1)$

Answer 1

1+9+25=2n-1
35=2n-1
2n-1-35
2n-36
n-36/2
n-18

Answer 2

Answer:

1² + 3² + 5² +.....................+(2n-1)² = (n/3)(2n-1)(2n+1)

Step-by-step explanation:

1² + 3² + 5² +.....................+(2n-1)² = (n/3)(2n-1)(2n+1)

p(1) = (2*1 - 1)²  = (1/3)(2*1 - 1)(2*1 + 1)

=> 1² = (1/3)(1)(3)

=> 1 = 1

p(2) = 1² + (2*2 - 1)² = (2/3)(2*2-1)(2*2 + 1)

=> 1 + 3² = (2/3)(3)(5)

=> 10 = 10

let assume

p(k) = 1² + 3² + 5² +.....................+(2k-1)² = (k/3)(2k-1)(2k+1)

then

p(k + 1) =

1² + 3² + 5² +.....................+(2k-1)² + (2(k+1) - 1)²

=  (k/3)(2k-1)(2k+1) + (2k + 1)²

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

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

= (2k + 1)  ( 2k² - k + 6k + 3) /3

= (2k + 1)  ( 2k²  + 5k + 3) /3

= (2k + 1)  ( 2k²  + 2k + 3k + 3) /3

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

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

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

= RHS

Hence

1² + 3² + 5² +.....................+(2n-1)² = (n/3)(2n-1)(2n+1)