The sum of the cubes of any number of consecutive integer starting with 1 is the square of some integer.

1³ + 2³ = 3²

1³ + 2³ + 3³ = 6²

1³ + 2³ + 3³ + 4³ = 10²

1³ + 2³ + 3³ + 4³ +5³ =15²

1³ + 2³ + 3³ + 4³ + 5³ + 6³=21²

1³ + 2³ + 3³ + 4³ + 5³ +6³ + 7³= 28²

Deriving RHS:

3->6->10->15->21->28

The RHS gets increased in some pattern. Can you recognize the pattern ?

Many who are good in vedic mathematics can easily recognize this.

3+3=>6 ; 6+4=>10; 10+5=>15; 15+6=>21; 21+7=>28 and this goes on.

Note that, I am here talking about the RHS alone.

The next step of solution,

Let’s start with example

1³ + 2³ + 3³ + 4³ + 5³ +…………….+300³ = (n) ²

If we follow the above step we need the previous RHS value i.e., 1³ + ………….. +299³ RHS value.

The numbers that we have now is only LHS so there is no other go other than LHS

Solution from LHS=>

The RHS s are 3,6,10,15,21,28

^· 1³ + 2³ = 3²

Take last number from the series is 2 here and the RHS is 3

2+2*(0.5)=> 2 +1 => 3 =>RHS

^· 1³ + 2³ + 3 = 6²

Last number in the series is 3 and the RHS is 6

3 + 3*(1)=> 3 + 3 =>6 =>RHS

^· 1³ + 2³ + 3³ + 4³ = 10²

Last number in this series is 4 and the RHS is 10

4+ 4*(1.5) => 4 + 6 =>10 =>RHS

^· 1³ + 2³ + 3³ + 4³ +5³ =15²

Last number in this series is 5 and the RHS is 15

5+ 5*(2) => 5 + 10 =>15 =>RHS

^· 1³ + 2³ + 3³ + 4³ + 5³ + 6³=21²

Last number in this series is 6 and the RHS is 21

6+ 6*(2.5) => 6 + 15 =>21 =>RHS

Check out the bracketed values () while finding the RHS of the above series

In gets increased in a pattern that is from 0.5 to 1 to 1.5 to 2 to 2.5 and so on

So we write it in general for all the series

1³+2³ + 3³ + 4³ + 5³ +……………………….. +x³ = n²

X+ X *(the increment value for eg., 6.5) =n

The other problem that we get here is we must know the previous increment value then only we can add 0.5 to that and get the new one.

Finding the increment Value:

• 1³ + 2³ = 3² the increment value for this is 0.5. 2+2*(0.5)=>3=>RHS

2 + 2 * (2/2 -0.5) = 2+ 2 * (1-0.5) =2 + 2 *(0.5)

• 1³ + 2³ + 3³ = 6² the increment value for this is 1 . 3+3*(1)=>6=>RHS

3 + 3 * (3/2 -0.5) = 3+ 3 * (1.5-0.5) =3 + 3 *(1)

• 1³ + 2³ + 3³+ 4³ = 10² the increment value for this is 1.5. 4+4*(1.5)=>10=>RHS

4 + 4 * (4/2 -0.5) = 4+ 4 * (2-0.5) =4 + 4 *(1.5)

• 1³ + 2³ + 3³ +4³ + 5³= 15² the increment value for this is 2 . 5+5*(2)=>15=>RHS

5 + 5 * (5/2 -0.5) = 5+ 5 * (2-0.5) =5 + 5 *(1)

In general to find the increment we write as

([Last number in the series /2] -0.5)

Finally the general solution for the series is

1³+2³ + 3³ + 4³ + 5³ +……………………….. +x³ = [ x + x * ([x/2] -0.5)]²