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

Question

Swapnil Saxena · Answer

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

Ketan Chandak · Answer

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....)

Ashwin Muralidharan IIT Madras · Answer

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).

Sathya · Answer

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 , =

Sathya · Answer

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

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