Can it be convergent?

Replies

• 

  Morningdot Hablu

  The equation of nth term of this series is (1+2n)/4((n^2-n+1))^2;
  the numerator of the first term and last terms are in AP so you can easily found the numbers of terms.
  The number of terms is 9.
  So integrate the nth term equation from 1 to 9 and you got the answer.
  The answer is 1.4999.
  .
  Here i show hoe to get the nth term equation.
  3/(2^2) +5/(6^2) +7/(12^2) +..........17/(72^2) +19/(90^2);
  3/((2*1)^2) +5/((2*3)^2) +7/((2*6)^2) +..........17/((2*36)^2) +19/((2*45)^2);
  now from denumerator select the variables.
  1,3,6,............36,45.
  now the diff is
  2,3..........9.
  take the c.difference of 1,3,6......series is n and then try to find out the equation you get the equation as mentioned above.
  as we see that the numerators are in AP so you can easily get the numerator equation.
• 

  Sachin Jain

  @ freak16
  Very nice question..
  revised my lots of cncepts.
  
  @MohitKumar
  Your approach is very right. Infact i got the solution using your approach. Nice brainy work but i think you have made a little mistake in calculating the nth term.
  See here it goes :-
  
  For Numerator :- 3,5,7,... So 2n+1
  For denaominator :-
  4,36,144,....,8100
  Take 4 common :- 4(1^2,3^2,6^2.......45^2)
  Or we can write 4 ([1]sq , [1+2]sq, [1+2+3]sq ,......[1+2...+9]sq)
  
  So general term for denominator will be :-
  4(n^2(n+1)^2)/4
  Now 4 gets cancelled out
  We get the whole nth term as
  
  Tn=2n+1/(n^2(n+1)^2)
  
  Now 2n+1=(n+1)^2-n^2
  And we get
  Tn=1/(n+1)^2 - 1/n^2
  
  Now Put the values to find the summation and the values gets cancelled out (Simple Vn method)
  And we are left with
  Sum=1/1^2 minus 1/10^2
  Hence answer is 0.99
  Option (a) is correct
• 

  freak16

  @Blunderboy
  Hats off to you....😁
  you completely rock!!
  Exactly..
• 

  Deepika Bansal

  @blunder boy: please tell me about Vn method....
  these type of questions are a bit difficult for me to understand... Do they have some shortcut or some easier way..??
• 

  Morningdot Hablu

  Deepika Bansal

  @blunder boy: please tell me about Vn method....

  Put the value of n in nth term one by one and do the summation of all the terms you get the result.
  @BB nice find !!
  Actually i am in hurry so may be i have done some mistake there but ya the method is correct.
• 

  Deepika Bansal

  Thank you mohit....
• 

  Sachin Jain

  @ Deepika
  Sorry for being late.
  Well Mohit has done the job.
  But if you want to understand the method more in depth, you can refer to RD Sharma.
  You will find it there.
  Still i would like to give just a gist of that.
  Suppose we have to find summation of 1/1.2 + 1/2.3 + 1/3.4 +.....1/99.100 Here .(dot) signifies multiply 3.4 means 3 * 4
  Step 1 -> Find the nth term
  In our case it would be 1/n(n+1)
  Step 2-> Break it in partial fractions
  So it becomes
  Tn = 1/n - 1/(n+1)
  Step 3 -> Since we have to find summation that means T1+T2+T3+.....Tn
  So
  T1=1/1-1/2
  T2=1/2-1/3
  .
  .
  T98=1/98-1/99
  In our case n is 99
  T99=1/99-1/100
  
  Add all these and you will notice that 2nd part of any term gets cancelled by 1st part of next term. So we are left with
  Sum=1/1-1/100
  Sum=99/100
  Sum=0.99
  
  I hope its clear now....
• 

  Deepika Bansal

  Thank you blunder boy........