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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
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