What is the formula to find sum of cube of 1>odd natural numbers 2> even natural numbers?

Question

Maths + formula

Answer 1

we know odd number = (2n +1) n = 0 to infinity

(2n+1)^3 = 8n^3+ 12n^2 + 18n + 1

now summation = sum ( 8k^3+ 12k^2 + 18k +1) k from 0 to (n-1)/2

for even numbers it is sum (8n^3) sum k from 0 to n/2

you can plug in for sum of numbers, squares and cubes and simplify

math kp

Answer 2

Greetings,

sum of cubes natural numbers from 1 to n is (n(n+1)/2)^2 ...1)

sum of cubes of n even natural numbers is

2^3 + 4^3 + ... + (2n)^3

= 8(1^3 + 2^3 + ... + n^3)

= 8(n(n+1)/2)^2

= 2(n(n+1))^2 ...2)

sum of cubes of n odd natural numbers is the difference of

1^3 + 3^3 + ... + (2n - 1)^3

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

- ( 2^3 + 4^3 + ... + (2n)^2)

= ((2n)(2n + 1)/2)^2 - 2(n(n + 1))^2

= n^2(2n^2 - 1)

Regards