Product of the cube roots of two consecutive cubes is odd or even
Answers
Step-by-step explanation:
We know that sum of cubes of first n natural numbers is = (n(n+1)/2)2.
Sum of cube of first n even natural numbers
23 + 43 + 63 + ……… + (2n)3
Even Sum = 23 + 43 + 63 + .... + (2n)3
if we multiply by 23 then
= 23 x (13 + 23 + 32 + .... + (n)3)
= 23 + 43 + 63 + ......... + (2n)3
= 23 (n(n+1)/2)2
= 8(n(n+1))2/4
= 2(n(n+1))2
Example :
Sum of cube of first 4 even numbers = 23 + 43 + 63 + 83
put n = 4 = 2(n(n+1))2
= 2*(4*(4+1))2
= 2(4*5)2
= 2(20)2
= 800
8 + 64 + 256 + 512 = 800
Program for Sum of cubes of first n even numbers
Sum of cube of first n odd natural numbers
We need to compute 13 + 33 + 53 + …. + (2n-1)3
Answer:
The product of the cube roots of two consecutive cubes is odd