Flag Algebra> Sum of series...
question mark

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

Vishal Kamalakannan , 13 Years ago
Grade 11
anser 5 Answers
Swapnil Saxena

Last Activity: 13 Years ago

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

Ketan Chandak

Last Activity: 13 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

Last Activity: 13 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

Last Activity: 13 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

Last Activity: 13 Years ago

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

 

Done

Provide a better Answer & Earn Cool Goodies

Enter text here...
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free

Enter text here...