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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 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


 


 

Grade:11

5 Answers

Swapnil Saxena
102 Points
9 years ago

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

Ketan Chandak
15 Points
9 years ago

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
290 Points
9 years ago

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
35 Points
9 years ago

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
35 Points
9 years ago

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

 

Done

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