Question

Cubes of natural numbers are grouped as 1^3,(2^3,3^3),4^3,5^3,6^3).....Then the sum of numbers in nth group is

Answer 1

Answer:

Step-by-step explanation:

As seen the first group contains 1 element and the second group contains 2 elements and so on. Thus, the nth group will contain n elements.

Now, till first group 1 number is already grouped, till second group (1+2=)3 numbers are already grouped and till third group (1+2+3=)6 numbers are already grouped.Thus, till nth group (1+2+3+4....n=)n*(n+1)/2 numbers will be already grouped.So, the nth group will be like :

{((n*(n+1)/2)+1)^3,((n*(n+1)/2)+2)^3,((n*(n+1)/2)+3)^3...............((n*(n+1)/2)+n)^3}

Now, sum of  cube of first n natural numbers will be

((n*(n+1))/2)^2

So , the required sum will be= sum of cubes of first ((n*(n+1)/2)+n) natural numbers-sum of cubes of  first (n*(n+1)/2) natural numbers