The series is:

1^3+3*(2^2)+3^3+(3*(4^2))+5^3..................................

upto n terms(n is odd). Find the sum upto n terms.

I tried to solve the problem on the following lines. I separated the series into two. Sum of cubes and sum of squares. I found out the individual terms of series also. My doubt is the last term of those individual series. Please help me

The n th term of the first series will be (2n-1)³ and (2n)² respectively

given that n is odd last term of the series will be n^3....
so u will have to sum two series :

1^3+3^3 ...... +n^3(note here n is the no of terms of the main series and not this sub-series....)

and another is

3(2^2+4^2.......+(n-1)^2)   (here too n is as mentioned above....)

Hi Vishal,

Let n = 2k+1, where k is any natural number so that n is always odd.

Now series is S = 1³+3³+5³+.....(2k+1)³ + 3*{2²+4²+......(2k)²}.

Now sum this, and substitute back for k = (n-1)/2 in the final result.

Regards,

Ashwin (IIT Madras).

Given:1^3 + 3(2^2) + 3^3 + 3(4^2) + ...... + (2n-1)^3 + 3(4n^2).

Split this into 2:

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

and 3[2^2 + 4^2 + ... + 2n^2].

Use basic formulas,

And u would get the answer as ,

=

Answer is (n+1)(2n^3+2n-1)

Done